Parse error: syntax error, unexpected '"' in ....

**Scooby10** · Jan 18 '10, 04:17 AM

Switch your quotes around...

**riverdale1567** · Jan 18 '10, 05:13 AM

Bigger Question...

First thanks Scooby for your help.
I am trying to post data from 2 different scripts to the query you see below

Code:

$query = 'SELECT * FROM BIZ_APARTMENTS WHERE ((bizState="' . mysql_real_escape_string($_POST['b']) .
'"&&(bizCity="' . mysql_real_escape_string($_POST['c']) . '")';

the post data is coming from 2 different pages.

This one generates a drop down

Code:

<?php
/*  Program name: buildSelect.php
 *  Description:  Program builds a selection list
 *                from the database.
 */
?>
<html>
<head><title>Building info by state</title></head>
<body>
<?php
  $user="XXXXXXXXXXt";
  $host="XXXXXXXXXXXX";
  $password="XXXXXXXXX";
  $database = "XXXXXXXXXXXX";

  $cxn = mysqli_connect($host,$user,$password,$database)
         or die ("couldn't connect to server");
  $query = "SELECT DISTINCT bizState FROM BIZ_APARTMENTS ORDER BY bizState";
  $result = mysqli_query($cxn,$query)
            or die ("Couldn't execute query.");

 /* create form containing selection list */
  echo "<form action='processform41.php' method='POST'>
        <select name='b'>\n";

  while ($row = mysqli_fetch_assoc($result))
  {
     extract($row);
     echo "<option value='$bizState'>$bizState\n";
  }
  echo "</select>\n";
  echo "<input type='submit' value='Select State in which building is located'>
        </form>\n";
?>
</body></html>

and passes the info post [b] to this script

Code:

<?php

$username="XXXXXXXXX";
$password="XXXXXXXXXX";
$database="XXXXXXXXXXt";
$table="BIZ_APARTMENTS";
$column="bizState";

error_reporting(E_ALL);
mysql_connect("localhost",$username,$password);
@mysql_select_db($database) or die( "Unable to select database");


$query = "SELECT DISTINCT bizCity FROM BIZ_APARTMENTS WHERE bizState='" . mysql_real_escape_string($_POST['b']) . "'";


$result=mysql_query($query);
$ret = mysql_query($query) or die(mysql_error());
$num=mysql_numrows($result);






 echo "<b><center>Cities in State</center></b><br><br>";

 /* create form containing selection list */
  echo "<form action='city.php' method='POST'>
        <select name='c'>\n";

  $i=0;
  while ($i < $num) {


  $city=mysql_result($result,$i,"bizCity");


     echo "<option value='$city'>$city\n";
     $i++;
  }
  echo "</select>\n";
  echo "<input type='submit' value='Select City in which building is located'>
        </form>\n";
 print_r($result);

 ?>

which in turn should pass both post[b] and post[c] to this form for output

Code:

<?
$username="xxxxxxxxxx";
$password="xxxxxxxxxxx";
$database="xxxxxxxxx";
$table="xxxxxxxxx";
$column="bizState";
error_reporting(E_ALL);

mysql_connect("localhost",$username,$password);
@mysql_select_db($database) or die( "Unable to select database");




$query = 'SELECT * FROM BIZ_APARTMENTS WHERE ((bizState="' . mysql_real_escape_string($_POST['b']) .
'"&&(bizCity="' . mysql_real_escape_string($_POST['c']) . '")';



$result=mysql_query($query);
$ret = mysql_query($query) or die(mysql_error());
$num=mysql_numrows($result);

  mysql_close();




 echo "<b><center>Buildings in City</center></b><br><br>";

$i=0;
while ($i < $num) {
$name=mysql_result($result,$i,"bizName");
$address=mysql_result($result,$i,"bizAddr");
$city=mysql_result($result,$i,"bizCity");
$state=mysql_result($result,$i,"bizState");
$zip=mysql_result($result,$i,"bizZip");
$phone=mysql_result($result,$i,"bizPhone");
$email=mysql_result($result,$i,"bizEmail");

 echo "<b>Name: $name</b><br>Phone: $phone<br>Type: $type<br>Address: $address<br>City: $city<br>State: $state<br>Zip: $zip<br>Email:$email<br>";

$i++;
 }
print_r($city);
 ?>

BUT I am getting error message

Notice: Undefined index: b in /home/attorney/public_html/in-url.com/bigdump/city.php on line 15
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

thanks again, the help i have received from this forum has been priceless..

**Scooby10** · Jan 18 '10, 05:23 AM

Could it be the && in the query? Try replacing that with AND..

**Dormilich** · Jan 18 '10, 06:29 AM

that’s not it, there’s simply no field named "b" in the last script.

**riverdale1567** · Jan 18 '10, 10:34 AM

so how do I transfer data from the field [b] in the first script into the 3rd script so the query in the 3rd script can use fields 'b' and 'c'?

Code:

$query = 'SELECT * FROM BIZ_APARTMENTS WHERE ((bizState="' . mysql_real_escape_string($_POST['b']) .
'"&&(bizCity="' . mysql_real_escape_string($_POST['c']) . '")';

**Dormilich** · Jan 18 '10, 10:58 AM

you could add a hidden field to pass down the b value

Code:

echo '<input type="hidden" name="b" value="' . $_POST["b"] . '">';

**riverdale1567** · Jan 18 '10, 11:55 AM

syntax error in line 1...

First, let me say thank you again to both of you for you time and effort in helping me.
In my 3rd script, I am still getting the error:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

this is the script:

Code:

<?
$username="xxxxxxxxxx";
$password="xxxxxxxxx";
$database="xxxxxxxxxx";
$table="BIZ_APARTMENTS";
$column="bizState";


mysql_connect("localhost",$username,$password);
@mysql_select_db($database) or die( "Unable to select database");




$query = 'SELECT * FROM BIZ_APARTMENTS WHERE ((bizState="' . mysql_real_escape_string($_POST['b']) .
'"&&(bizCity="' . mysql_real_escape_string($_POST['c']) . '")';



$result=mysql_query($query);
$ret = mysql_query($query) or die(mysql_error());
$num=mysql_numrows($result);

  mysql_close();




 echo "<b><center>Buildings in City</center></b><br><br>";

$i=0;
while ($i < $num) {
$name=mysql_result($result,$i,"bizName");
$address=mysql_result($result,$i,"bizAddr");
$city=mysql_result($result,$i,"bizCity");
$state=mysql_result($result,$i,"bizState");
$zip=mysql_result($result,$i,"bizZip");
$phone=mysql_result($result,$i,"bizPhone");
$email=mysql_result($result,$i,"bizEmail");

 echo "<b>Name: $name</b><br>Phone: $phone<br>Type: $type<br>Address: $address<br>City: $city<br>State: $state<br>Zip: $zip<br>Email:$email<br>";

$i++;
 }
print_r($city);
 ?>

Also I am not getting any output either.

Thank you very much,
Josh

**Dormilich** · Jan 18 '10, 12:16 PM

there are 2 closing parentheses missing in the SQL.

**riverdale1567** · Jan 18 '10, 12:25 PM

Dormilich
Dude you are awesome, thanks ever so much. It always amazes when the s*** actually works.
Danke

**Dormilich** · Jan 18 '10, 12:27 PM

thinking like a parser helps. and experience.

**riverdale1567** · Jan 18 '10, 12:30 PM

Sorry about the profanity, it the Bronx in me that just comes out sometimes :-)

**Dormilich** · Jan 18 '10, 12:32 PM

that’s why there is an "edit" link.

**riverdale1567** · Jan 18 '10, 12:35 PM

How many years have you been programming?

**Dormilich** · Jan 18 '10, 12:41 PM

started with HTML and JS about 10 years ago and doing PHP for about 5 years. but that’s not much since I do it for a hobby.

Parse error: syntax error, unexpected '"' in ....

Parse error: syntax error, unexpected '"' in ....

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