Html form

**Ciary** · May 15 '09, 02:00 PM

first: set the form method to post like below:

Code:

<form id="aform" method="post">

then, check if you did a post

Code:

if($_SERVER['REQUEST_METHOD'] == 'post'){

after that you simply use an insert query

Code:

$conn = mysql_connect("myDB","root","");
$result = mysql_query("INSERT INTO myTable (field1, field2) VALUES ('".$_POST['field1']."', '".$_POST['field2']."')");

if you have multiple tables you need more then one insert-query

**Dormilich** · May 15 '09, 08:05 PM

Originally posted by Ciary

then, check if you did a post

Code:

if($_SERVER['REQUEST_METHOD'] == 'post'){

another way to check if something was posted:

Code:

// HTML
<input type="submit" name="submit" value="submit form">

// PHP
if (isset($_POST['submit'])) { … }

**Atli** · May 15 '09, 08:39 PM

Originally posted by Dormilich

another way to check if something was posted:

Code:

// HTML
<input type="submit" name="submit" value="submit form">

// PHP
if (isset($_POST['submit'])) { … }

There is one cavity in that.
If the form is submitted without the use of the submit button (by pressing Enter, for example), the submit button will not be included in the request.

A way around this is to include a hidden element and use that to check.

Code:

// HTML
<input type="hidden" name="submitted" value="1" />
<input type="submit" name="submit" />

// PHP
if(isset($_POST['submitted'])) {
  ...
}

Originally posted by Ciary

[...]
after that you simply use an insert query

Code:

$conn = mysql_connect("myDB","root","");
$result = mysql_query("INSERT INTO myTable (field1, field2) VALUES ('".$_POST['field1']."', '".$_POST['field2']."')");

if you have multiple tables you need more then one insert-query

Just make sure you don't actually put the $_POST fields directly into the query, or your database will be vulnerable to SQL Injection.

**Dormilich** · May 15 '09, 09:53 PM

Originally posted by Atli

There is one cavity in that.
If the form is submitted without the use of the submit button (by pressing Enter, for example), the submit button will not be included in the request.

geez, you never stop learning new things here… I'll remember that.

**luckysanj** · May 16 '09, 07:12 AM

Thank you for your all cooperation & it is very helpful for me. But i have not got my actually answer. It may be i was not able to clear my problem with you.

So once more i am trying to clear you my problem.
Dear sir, I have three different database table like:

tbl_a, tbl_b, tbl_c & it has different field also.

& I want to insert data in these different table through one html form.

so plz guide me is it possible to insert data in different database table through on html form.

**Markus** · May 16 '09, 07:34 AM

Multiple inserts, through one query, can be achieved with syntax like:

Code:

INSERT INTO tbl_name (a,b,c) VALUES(1,2,3),(4,5,6),(7,8,9);

Not sure what the earliest version of MySQL this will work with, but it does for 5.1.

**luckysanj** · May 16 '09, 08:12 AM

It doesn't work my mysql cuz i have mysql 5.

give me another idea or method to input data through one html form to different database table.

**Markus** · May 16 '09, 08:39 AM

Originally posted by luckysanj

It doesn't work my mysql cuz i have mysql 5.

give me another idea or method to input data through one html form to different database table.

Run separate INSERT queries for each table.

**luckysanj** · May 17 '09, 07:58 AM

Hello Ciary, Can u give me idea to insert data into multiple table like tbl_a, tbl_b, tbl_c through one HTML form.
there is::
3 text box in single form, linked to 3 diffrent tables,
do u have any idea to insert data to 3 different tables.

**Dormilich** · May 17 '09, 08:26 AM

you just nned to know which input (aka which POST index) goes to which database and then construct the SQL accordingly.

**Ciary** · May 17 '09, 10:18 AM

an example:

Code:

<?php
if($_SERVER['REQUEST_METHOD'] == 'post'){
$conn = mysql_connect("myDB","root","");

$result = mysql_query("INSERT INTO myfirsttable (field1,field2,field3) VALUES (".$_POST('myfield1').",".$_POST('myfield2').",".$_POST('myfield3').")");

$result = mysql_query("INSERT INTO mysecondtable (field1,field2,field3) VALUES (".$_POST('myfield4').",".$_POST('myfield5').",".$_POST('myfield6').")");

$result = mysql_query("INSERT INTO myfirsttable (field1,field2,field3) VALUES (".$_POST('myfield1').",".$_POST('myfield3').",".$_POST('myfield5').")");

mysql_close($conn);
}
?>

<form method=post>
<input type='text' id='field1' name='field1' />
<input type='text' id='field2' name='field2' />
<input type='text' id='field3' name='field3' />
<input type='text' id='field4' name='field4' />
<input type='text' id='field5' name='field5' />
<input type='text' id='field6' name='field6' />
</form>

this is an example with multiple values inserted in multiple tables. normally (as atli discribed) you don't put post-variables directly in your query to avoid SQL injection.

**luckysanj** · May 17 '09, 03:14 PM

IN UR GUIDENCE I have GOT NEW IDEA & THANK YOU VERY MUCH for ur cooperation.

IN

$result = mysql_query("IN SERT INTO myfirsttable (field1,field2, field3) VALUES (".$_POST('myfiel d1').",".$_POST('myfiel d2').",".$_POST('myfiel d3').")");

In these above underline line there is error. So i use little bit change on ur method.

Now it works very well. & can u tell me below this my method is right or not.

Code:

<?php
foreach($_POST as $key=>$value)
	{
	$$key=$value;
	}
$link=mysql_connect("localhost","root","") or die("Connection Failed");
mysql_select_db("test",$link) or die("Database Not Found");

mysql_query("INSERT INTO tbl_personal (per_name,per_address) VALUES ('$per_name','$per_address')");
mysql_query("INSERT INTO tbl_employment (org_name,org_address) VALUES ('$org_name','$org_address')");
mysql_query("INSERT INTO tbl_technical (tec_course,tec_institution) VALUES ('$tec_course','$tec_institution')");
mysql_close($link);

?>
<body>
<form id="form1" name="form1" method="post" action="">
  <table width="71%" border="1" align="center">
    <tr>
      <td width="48%"><strong>Personal Details </strong></td>
      <td width="52%"><strong>Employement Details </strong></td>
    </tr>
    <tr>
      <td><table width="100%" border="0">
        <tr>
          <td width="35%">Name:</td>
          <td width="65%"><input name="per_name" type="text" id="per_name" /></td>
        </tr>
        <tr>
          <td>Address:</td>
          <td><input name="per_address" type="text" id="per_address" /></td>
        </tr>
        
      </table></td>
      <td><table width="100%" border="0">
        <tr>
          <td width="47%">Organization Name:</td>
          <td width="53%"><input name="org_name" type="text" id="org_name" /></td>
        </tr>
        <tr>
          <td>Address</td>
          <td><input name="org_address" type="text" id="org_address" /></td>
        </tr>
        
      </table></td>
    </tr>
    <tr>
      <td><strong>Technical Qualification </strong></td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td><table width="100%" border="0">
        <tr>
          <td width="34%">Course Covered: </td>
          <td width="66%"><input name="tec_course" type="text" id="tec_course" /></td>
        </tr>
        <tr>
          <td>Institution</td>
          <td><input name="tec_institution" type="text" id="tec_institution" /></td>
        </tr>
        
      </table></td>
      <td>&nbsp;</td>
    </tr>
    <tr>
      <td><input name="submit" type="submit" id="submit" value="Submit" /></td>
      <td>&nbsp;</td>
    </tr>
  </table>
</form>

**Ciary** · May 17 '09, 04:28 PM

Code:

# <?php
# if($_SERVER['REQUEST_METHOD'] == 'post'){
# $conn = mysql_connect("myDB","root","");
#  
# $result = mysql_query("INSERT INTO myfirsttable (field1,field2,field3) VALUES (".$_POST('myfield1').",".$_POST('myfield2').",".$_POST('myfield3').")");
#  
# $result = mysql_query("INSERT INTO mysecondtable (field1,field2,field3) VALUES (".$_POST('myfield4').",".$_POST('myfield5').",".$_POST('myfield6').")");
#  
# $result = mysql_query("INSERT INTO mythirdtable (field1,field2,field3) VALUES (".$_POST('myfield1').",".$_POST('myfield3').",".$_POST('myfield5').")");
#  
# mysql_close($conn);
# }
# ?>
#  
# <form method=post>
# <input type='text' id='myfield1' name='myfield1' />
# <input type='text' id='myfield2' name='myfield2' />
# <input type='text' id='myfield3' name='myfield3' />
# <input type='text' id='myfield4' name='myfield4' />
# <input type='text' id='myield5' name='myfield5' />
# <input type='text' id='myfield6' name='myfield6' />
# </form>

srr there was a small typo in my code here is the right version (without errors)

i guess your code is correct.it would be easier to read if you used code tags.

**luckysanj** · May 17 '09, 07:07 PM

Still ur same code doesn't work. But it's ok i have solve my problem with ur good guidance So thank you very much.

Html form

Html form

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