Html form

[luckysanj]

Through one html form how to insert data into different database table.

[Ciary]

first: set the form method to post like below:

<form id="aform" method="post">

then, check if you did a post

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

after that you simply use an insert query

$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]

@Ciary
another way to check if something was posted:

// HTML
<input type="submit" name="submit" value="submit form">
 
// PHP
if (isset($_POST['submit'])) { … }

[Atli]

@Dormilich
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.

// HTML
<input type="hidden" name="submitted" value="1" />
<input type="submit" name="submit" />
 
// PHP
if(isset($_POST['submitted'])) {
  ...
}

@Ciary
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]

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

[luckysanj]

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]

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

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]

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]

@luckysanj
Run separate INSERT queries for each table.

[luckysanj]

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]

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

[Ciary]

an example:

<?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]

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

IN

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

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.

<?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]

# <?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]

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.

[Dormilich]

just for info: $_POST['key'] (with square brackets, but that's probably a typo)

[luckysanj]

thank you, Dormilich for replying. It is now working.

Is there any function on php which reads the last value of data on the database table.
For eg: On database table
tbl_a

std_id name
1 Ram
2 Shyam
3 Hari
4 Kishor

In these above table. I want to read that last value of std_id(4) through php function or any idea if u have .

& one thing more is How to use primary key & foreign key on sql database.

[Ciary]

seems like there were a lot of typo's in my code. i also forgot mysql_select_db but you already knew i did.

the request is very simple i think:

$result = mysql_query("SELECT * FROM tbl_a");
$anArray = mysql_fetch_assoc($result);
 
$lastRecord = $anArray[(count($anArray)-1)];
 

for the question about primary key and foreign key, i would refer to the mysql forum since it isnt php anymore and ... i can't really answer that :-O

[luckysanj]

It doesn't work ciary sir.

[Dormilich]

that's because only the first row was fetched…

depending on your DB system, there are functions that will fetch all results in one go (like PDOStatement->fetchAll()) where this would work.

[Markus]

Use ORDER BY on std_id and then LIMIT 1.

SELECT * FROM `tbl_name` ORDER BY `std_id` DESC
 

[luckysanj]

Thank you .

[Canabeez]

@luckysanj
Hi, using this code won't protect you from SQL Injection, you're still inserting direct $_POST vars into your SQL query, I suggest you at least do something like:

$$key = addslashes($value);
/* or */
$$key=mysql_real_escape_string($value);

And to be more accurate, the query for multiple inserts, as Markus mentioned

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

Expand|Select|Wrap|Line Numbers
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.

that is available since mySQL 4, so it should work perfectly (in case you are using mySQL 4 or higher), otherwise you must probably have sql syntax error or your number of inserts is limited by configuration.

[luckysanj]

Thank you very much .