search engine cant function

**Motoma** · Jan 31 '07, 03:01 PM

Originally posted by bb nicole

Below is my search engine for job portal which jobseeker can find the job through quick search. But it cant work... Is it mysql query got problem?? Thanx..

Interface
[HTML]<html>
<head>
<title>UMS e-Job Portal</title>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
<style type="text/css">

</style></head>

<body bgcolor="#FFFFF F" text="#000000">
<form name="search job" method="post" action="jobsear ch_act.php">
<p> </p>

<table width="600" border="0">
<tr>
<td colspan="3"><fo nt color="#000000" ><strong>QUIC K SEARCH</strong></font></td>
</tr>
<tr>
<td width="217"><fo nt color="#000000" >Keyword(s)</font><font color="#000000" > </font></td>
<td width="30"><div align="center"> <strong>:</strong></div></td>
<td width="339"><in put name="keyword" type="text" maxlength="30"> </td>
</tr>
<tr>
<td><font color="#000000" >Job Category</font> </td>
<td><div align="center"> <strong>:</strong></div></td>
<td><select name="jobCatego ry">
<option selected></option>
<option>Account ing</option>
<option>Adverti sing</option>
<option>Agricul ture</option>
<option>Banking </option>
<option>Biotech nology</option>
<option>Chemica l Eng</option>
<option>Chemist ry</option>
<option>Clerica l/Admin</option>
<option>Custome r Service</option>
<option>Educati on</option>
<option>Electro nics Eng</option>
<option>Food Tech</option>
<option>Gener al Workers</option>
<option>Graphic s</option>
<option>Healthc are</option>
<option>Hotel/Food</option>
<option>Human Resources</option>
<option>IT/Computer</option>
<option>Journal ist/Editor</option>
<option>Manufac turing</option>
<option>Marketi ng</option>
<option>Mechani cal Eng</option>
<option>Media/Entertain</option>
<option>Persona l Care</option>
<option>Publish ing</option>
<option>Purchas ing</option>
<option>Quantit y Survey</option>
<option>Sales </option>
<option>Scien ce/Tech</option>
<option>Secreta rial</option>
<option>Socia l Services</option>
<option>Trainin g</option>
<option>Other s</option>
</select></td>
</tr>
<tr>
<td><font color="#000000" >Job Location</font></td>
<td><div align="center"> <strong>:</strong></div></td>
<td><select name="jobLocati on">
<option selected></option>
<option>Perli s</option>
<option>Penan g</option>
<option>Kedah </option>
<option>Perak </option>
<option>Kelanta n</option>
<option>Terengg anu</option>
<option>Pahan g</option>
<option>Selango r</option>
<option>Kuala Lumpur</option>
<option>Neger i Sembilan</option>
<option>Melak a</option>
<option>Johor </option>
<option>Labua n</option>
<option>Sabah </option>
<option>Sarawak </option>
</select></td>
</tr>
<td colspan="3"><di v align="center">
<input type=hidden name=act value=adduser>

<input name="Search" type="submit" id="Search" value="Search">
</div></td>
</table>
</form>
<p><img src="image/bottom%20banner .jpg" width="955" height="46"></p>
</body>
</html>[/HTML]

PHP CODE for search engine
[PHP]<?php
//required file for database connection
require("databa se.php");
if (isset($_POST['keyword'])) {
$keyword = mysql_real_esca pe_string($_POS T["keyword"]);
$category=mysql _real_escape_st ring($_POST["jobCategor y"]);
$location=mysql _real_escape_st ring($_POST["jobLocatio n"]);

echo"<p align=center><c enter><font face='Arial' size='3'>Search Results</p>";
//An error given when they did not enter a keyword.
if ($keyword== "") {
echo "<p align=center><f ont face='Arial' size='2'>Opps.. You forgot enter a keyword(s)";
echo"<br><br><b r><a href='jobsearch .php? '><font face='Arial' size='2'>Back</a>";

exit;
}

$sql="SELECT * FROM job WHERE jobCategory like '$jobCategory' AND jobLocation like '$jobLocation' AND jobTitle LIKE '%$keyword%' OR jobType LIKE '%$keyword%'OR jobPosition LIKE'%$keyword% '";

$anymatches = @mysql_num_rows ($data);
if ($anymatches == 0)
{
echo "<center><f ont face='Arial' size='2'> Sorry, no job found.</center><br><br> ";
echo"<br><br><b r><a href='jobsearch .php? '><font face='Arial' size='2'>Back</a>";

exit();

}

$bgcolor="#f1f1 f1";
echo "<table border='1'align =center>";
echo"
<TABLE width=70% align=center cellpadding=0 cellspacing=0> <tr>

<td bgcolor='dfdfdf ' > <font face='arial' color='#000000' size='3'>Job Title</font></td>
<td bgcolor='dfdfdf ' > <font face='arial' color='#000000' size='3'>Compan y</font></td>
<td bgcolor='dfdfdf ' > <font face='arial' color='#000000' size='3'>Locati on</font></td></tr>
<td bgcolor='dfdfdf ' > <font face='arial' color='#000000' size='3'>Date Posted</font></td></tr>

</tr>";
$counter = 0;
while($result = @mysql_fetch_ar ray( $data )){
if ($counter == 3) {

$counter = 0;
}

if($bgcolor=='# f1f1f1'){$bgcol or='#ffffff';}
else{$bgcolor=' #f1f1f1';}

echo "<tr><td align=left bgcolor=$bgcolo r id='title'><a href='jobsearch _result.php?id= {$result['ID']}'><font face='Arial' size='2'>{$resu lt['jobTitle']}</a></td>
<td align=left bgcolor=$bgcolo r id='title'><fon t face='Arial' size='2'>{$resu lt['companyName']}</td>
<td align=left bgcolor=$bgcolo r id='title'><fon t face='Arial' size='2'>{$resu lt['jobLocation']}</td>
<td align=left bgcolor=$bgcolo r id='title'><fon t face='Arial' size='2'>{$resu lt['datePosted']}</td>
</tr>";
$counter = $counter + 1;
}

echo "</table>"; echo"<font face='Arial' size='2'>Keywor d: " .$keyword;
echo"<br><br><b r><a href='jobsearch .php? '><font face='Arial' size='2'>Back</a>";
}

?> [/PHP]

What doesn't work about this? Please elaborate. Do you get any errors appearing?

**bb nicole** · Feb 1 '07, 04:53 AM

When i insert the keyword, job category and job location, it will show:

Sorry, no job found.

The job is already in database, but the result is can't find the job, what is the problem??

Thanks..:)

**bb nicole** · Feb 1 '07, 09:13 AM

Could someone tell me what is the problem of my search engine.. I really can't find the problem.. Thanks..

**ak1dnar** · Feb 1 '07, 09:16 AM

To get a search result belongs to your search criteria you should have
at least one value in the JOB table for each an every input that
you passing from the form. why i am saying like that actually, your
sql query it contains with AND operators.

for a example see this: [This is your Query]

SELECT * FROM job WHERE jobCategory like 'IT/Computer'
AND jobLocation like 'Melaka' AND jobTitle LIKE '%programmer%'
OR jobType LIKE '%programmer%'O R jobPosition LIKE'%programme r%'

And also Note that %% are Missing.Try like this try to Print the $sql
value as i did here.
Then Manually try to Execute the values from the Table by using the SQL query window.
If you getting a out put you are ready to fly.
[PHP]
//Your Result Area
echo $sql;
//Your Result Area
[/PHP]

**bb nicole** · Feb 1 '07, 10:40 AM

my search engine is list menu, is it got simple ways to code the sql query?? Because is quick a lot sql query need to code..
Is it really need to code like this??

[PHP]
SELECT * FROM job WHERE jobCategory like 'IT/Computer'
AND jobLocation like 'Melaka' AND jobTitle LIKE '%programmer%'
OR jobType LIKE '%programmer%'O R jobPosition LIKE'%programme r%'
[/PHP]

**ak1dnar** · Feb 1 '07, 11:40 AM

I got this Nice Script from Ronald.
I have done modification for the script that suited for you.

before you start In HTML List menus do the modifaication like this for better performance of the script.
[HTML]

<option value="Others" >Others</option>
[/HTML]

No use of putting this "mysql_real_esc ape_string" for list menus.But put it to the text input box.

Here is the PHP:
Do the Modification as u wish.If u need you can pass AND instead of OR keyword here.If You passing or all the related data will list.

[PHP]
$keyword = mysql_real_esca pe_string($_POS T["keyword"]);
$category=$_POS T['jobCategory'];
$location=$_POS T["jobLocatio n"];

$string = array();
$where = "";
if (isset($_POST["keyword"]) AND !empty($_POST["keyword"]))
$string[] = " jobTitle LIKE '%".$_POST["keyword"]."%' ";
if (isset($_POST["keyword"]) AND !empty($_POST["keyword"]))
$string[] = " jobType LIKE '%".$_POST["keyword"]."%' ";
if (isset($_POST["keyword"]) AND !empty($_POST["keyword"]))
$string[] = " jobPosition LIKE '%".$_POST["keyword"]."%' ";
if (isset($_POST["jobCategor y"]) AND !empty($_POST["jobCategor y"]) AND ($category != Others))
$string[] = " jobCategory LIKE '%".$_POST["jobCategor y"]."%' ";
if (isset($_POST["jobLocatio n"]) AND !empty($_POST["jobLocatio n"]))
$string[] = " jobLocation LIKE '%".$_POST["jobLocatio n"]."%' ";
if (!empty($string ))
$where = " WHERE ".implode(" OR", $string);// OR/AND

$sql="SELECT * FROM job $where";
// Call for this $sql
[/PHP]

I couldn't test it against database, becoz i dont like to create tables for u and give the solution. Better to POST SQL table structures [Coding] also next time.

**bb nicole** · Feb 1 '07, 12:44 PM

Thanks for your help but i have copy the php code as you showed me, but it still can not work... The result show that no job found.. I really no idea for it.. My php code as below:
[PHP]
<?php
//required file for database connection
require("databa se.php");
if (isset($_POST['keyword'])) {
$keyword = mysql_real_esca pe_string($_POS T["keyword"]);
$jobCategory=$_ POST["jobCategor y"];
$jobLocation=$_ POST["jobLocatio n"];

$string = array();
$where = "";
if (isset($_POST["keyword"]) AND !empty($_POST["keyword"]))
$string[] = " jobTitle LIKE '%".$_POST["keyword"]."%' ";
if (isset($_POST["keyword"]) AND !empty($_POST["keyword"]))
$string[] = " jobType LIKE '%".$_POST["keyword"]."%' ";
if (isset($_POST["keyword"]) AND !empty($_POST["keyword"]))
$string[] = " jobPosition LIKE '%".$_POST["keyword"]."%' ";
if (isset($_POST["jobCategor y"]) AND !empty($_POST["jobCategor y"]) AND ($category != Others))
$string[] = " jobCategory LIKE '%".$_POST["jobCategor y"]."%' ";
if (isset($_POST["jobLocatio n"]) AND !empty($_POST["jobLocatio n"]))
$string[] = " jobLocation LIKE '%".$_POST["jobLocatio n"]."%' ";
if (!empty($string ))
$where = " WHERE ".implode(" OR", $string);// OR/AND

$sql="SELECT * FROM job $where";
// Call for this $sql

echo"<p align=center><c enter><font face='Arial' size='3'>Search Results</p>";
//An error given when they did not enter a keyword.
if ($keyword== ""||$jobCategor y==""||$jobLoca tion=="") {
echo "<p align=center><f ont face='Arial' size='2'>You are missing some required fields";
echo"<br><br><b r><a href='jobsearch .php? '><font face='Arial' size='2'>Back</a>";

exit;
}

$anymatches = @mysql_num_rows ($data);
if ($anymatches == 0)
{
echo "<center><f ont face='Arial' size='2'> Sorry, no job found.</center><br><br> ";
echo"<br><br><b r><a href='jobsearch .php? '><font face='Arial' size='2'>Back</a>";

exit();

}

$bgcolor="#f1f1 f1";
echo "<table border='1'align =center>";
echo"
<TABLE width=70% align=center cellpadding=0 cellspacing=0> <tr>

<td bgcolor='dfdfdf ' > <font face='arial' color='#000000' size='3'>Job Title</font></td>
<td bgcolor='dfdfdf ' > <font face='arial' color='#000000' size='3'>Compan y</font></td>
<td bgcolor='dfdfdf ' > <font face='arial' color='#000000' size='3'>Locati on</font></td></tr>
<td bgcolor='dfdfdf ' > <font face='arial' color='#000000' size='3'>Date Posted</font></td></tr>

</tr>";
$counter = 0;
while($result = @mysql_fetch_ar ray( $data )){
if ($counter == 3) {

$counter = 0;
}

if($bgcolor=='# f1f1f1'){$bgcol or='#ffffff';}
else{$bgcolor=' #f1f1f1';}

echo "<tr><td align=left bgcolor=$bgcolo r id='title'><a href='jobsearch _result.php?id= {$result['ID']}'><font face='Arial' size='2'>{$resu lt['jobTitle']}</a></td>
<td align=left bgcolor=$bgcolo r id='title'><fon t face='Arial' size='2'>{$resu lt['companyName']}</td>
<td align=left bgcolor=$bgcolo r id='title'><fon t face='Arial' size='2'>{$resu lt['jobLocation']}</td>
<td align=left bgcolor=$bgcolo r id='title'><fon t face='Arial' size='2'>{$resu lt['datePosted']}</td>
</tr>";
$counter = $counter + 1;
}

echo "</table>"; echo"<font face='Arial' size='2'>Keywor d: " .$keyword;
echo"<br><br><b r><a href='jobsearch .php? '><font face='Arial' size='2'>Back</a>";
}

?>
[/PHP]

This is my table and datatype
Tablename: job
job_ID int(10) PRI NULL auto_increment
company_ID int(10) 0
companyName varchar(30)
contactName varchar(30)
emailAdd varchar(50)
contactNum varchar(11)
contactAdd varchar(100)
jobTitle varchar(30)
jobType varchar(30)
jobCategory varchar(30)
jobPosition varchar(30)
jobLocation varchar(30)
mRequirement blob
jobDescription blob
datePosted date 0000-00-00

**bb nicole** · Feb 1 '07, 04:26 PM

What is the problem with my search engine, someone know what is the problem with my code??? Please help me to solve it....
Really thanks..

**Atli** · Feb 1 '07, 06:15 PM

Hi.

Do you ever query the database with the $sql you created?

If so I'm not seing it :P

**bb nicole** · Feb 1 '07, 06:51 PM

Sorry, i'm a newbie, i don't really understand what you mean...
Can u explain in more details...:)

**Atli** · Feb 1 '07, 07:04 PM

I see you create a query and use the mysql_num_row function.

What I dont see is where you acctually send the query to the database and get the results.

This might explain better

[PHP]// 1. Create query
$SQL = "SELECT * FROM MyTable";

// 2. Query database and get results
$RESULT = mysql_query($SQ L);

// 3. Count rows
echo "I have ". mysql_num_rows( $RESULT) ." rows in my table :D";[/PHP]

You have stepps 1 and 3 but I dont see stepp 2 anywhere

**Motoma** · Feb 1 '07, 07:05 PM

Originally posted by bb nicole

Sorry, i'm a newbie, i don't really understand what you mean...
Can u explain in more details...:)

In your code, you create an SQL query, and store it to the $sql variable, however, you never actually execute the SQL query in this code.

**bb nicole** · Feb 1 '07, 10:44 PM

I have add the php code regarding your suggestion, but it still display the message: No job found... And with the echo $sql; it will print the sql query in the page, but not call the database out... So, what should i do or what php code should i add again to make it work...

PHP Code
[PHP]
$sql="SELECT * FROM job $where";
// Call for this $sql

if(!empty($keyw ord)&&!empty($j obCategory)&&!e mpty($jobLocati on)){
$result = mysql_query($sq l);

echo $sql;
}
[/PHP]

**bb nicole** · Feb 2 '07, 10:08 AM

Is it anyone know what is my search engine problem??? Thanks...

search engine cant function

search engine cant function

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