Inserting Multiple rows in Mysql table with PHP

Question

I am in a fix here. I have code that does not insert multiple data into mysql table with one form. Here's my code-

<?php
if (isset($_POST['submitted'])) {
  include('config.php');
  foreach($_POST['name'] as $row=>$nam) {
    $name=$nam;
    $class=$_POST['class'][$row];
    echo "<br/>" . $name . "<br/>" . $class;
    $sql="INSERT INTO multiple (name, class) VALUES ('$name','$class')";
  }
  if (!mysql_query($sql)) die('Error: ' . mysql_error());
  echo "1 record added";
}
?>

<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
<label>Name1:
<input id="name" name="name[]" type="text">
<input type="hidden" name="submitted" value="true" />
</label>
<label> Class1:
<input id="class" name="class[]" type="text">
</label>
<label>Name2:
<input id="name" name="name[]" type="text">
</label>
<label>Class2:
<input id="class" name="class[]" type="text">
</label>
<label>Name3:
<input id="name" name="name[]" type="text">
</label>
<label>Class3:
<input id="class" name="class[]" type="text">
</label>
<label>Name4:
<input id="name" name="name[]" type="text">
</label>
<label>Class4:
<input id="class" name="class[]" type="text">
</label>
<label>Name5:
<input id="name" name="name[]" type="text">
</label>
<label>Class5:
<input id="class" name="class[]" type="text">
</label>
<input value="Add" type="submit">

</form>

When I press the submit button nothing inserts in the mysql table. Only empty fields are created. If I insert 5 text field I get 5 empty fields in sql table.

mysql_* is deprecated, and your code is open to SQL injection. — Waleed Khan
– Waleed Khan, Commented Aug 14, 2012 at 21:03
please show us the output of your echo statement. also echo $sql; — David Chan
– David Chan, Commented Aug 14, 2012 at 21:11
It's illegal to duplicate id's the way you are doing. Each value in id="" must only be used once. — Ariel
– Ariel, Commented Aug 14, 2012 at 21:13
Please show us the output of print_r($_POST); then we can debug this. Put that right at the top of the form. — Ariel
– Ariel, Commented Aug 14, 2012 at 21:13
1 record addedArray ( [name] => Array ( [0] => Arti [1] => [2] => [3] => [4] => ) [submitted] => true [class] => Array ( [0] => a [1] => [2] => [3] => [4] => ) ) this is shown while print_r($_POST); Here "Arti" is a Name and "a" is a class — Imroz Ahmad
– Imroz Ahmad, Commented Aug 14, 2012 at 21:19

chris · Accepted Answer · 2012-08-14 21:27:33Z

1

Imroz, your use of [] as part of the names of your input elements (not id's) example name="class[]" when the form is posted it builds an array. The post object PHP would recognize would be $_POST['class']

But that being an array means you have to handle it slightly different before inserting it into your database as you can't just (well maybe you can) drop an array into the DB

if you did

for($x=0;$x < count($_POST['class']); $x++)
{
    echo $_POST['class'][$x].'<br>';
}

you would be able to see all your posted inputs from the inputs with the name class[]

of course this is a core example of what you need to do overall, but I am just trying to express whats going on with your posted data.

answered Aug 14, 2012 at 21:27

chris

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3 Comments

Alex Kalicki Over a year ago

The OP's already taking into account that these are arrays using a foreach loop to iterate through.

Imroz Ahmad Over a year ago

Thank You Chris. I understand now, why empty fields are inserted. Thanks a lot Chris!

Imroz Ahmad Over a year ago

Chris, in the system you showed, is still inserting empty rows in the table :(

Pragnesh Chauhan · Accepted Answer · 2012-11-08 06:26:13Z

1

i think your inserting query is problem.....try like this

$query = mysql_query("INSERT INTO category VALUES('','$name','$class','')") or die(mysql_error());

edited Nov 8, 2012 at 6:26

Pragnesh Chauhan

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answered Nov 8, 2012 at 6:09

hussain

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Comments

Ariel · Accepted Answer · 2012-08-14 21:52:46Z

0

You have the } in the wrong place.

The } right after the $sql= should be moved after the echo "1 record added";

I reformatting your code to use proper indenting and the error was obvious.

answered Aug 14, 2012 at 21:52

Ariel

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