creating table in database dynamically

Question

I have this coding to create table in the database but the message($mess="Sucessfully Table Added";) is displayed but it is not creating table anywhere. any one could help. here are my code.

<?php 
if ($_POST["submit"]=="Add") {
$Sql="CREATE TABLE ".$_POST['title'] ."(";
for($i=0;$i<sizeof($_POST['column']);$i++)
{

$Sql.=$_POST['column'][$i] .$_POST['data_type'][$i] ."(".$_POST['size'][$i].")";
if($i!=(sizeof($_POST['column'])-1))
{
$Sql.=",";
}
}
$Sql.=")";
mysql_query($Sql);
$mess="Sucessfully Table Added";
}
?>
<form action="" method="post">
<div class="addTable">
 <span class="row_1">Add Table Name</span>
 <span class="row_2"><input type="text" class="inputbox" name="title" /></span>
<table id="myTable"></table>
<script>
$(document).ready(function () {
    $("#addRow").click(function () {
        $("#myTable").append("<div class='left title'>Column Title</div><div class='left'><input type='text' class='inputboxSmall' name='column[]'></div><div class='left title'>Data Type</div><div class='left'><input type='text' class='inputboxSmall' name='data_type[]'></div>&nbsp;&nbsp;<div class='left title'>size</div><div class='left clearfix'><input type='text' class='inputboxSmall' name='size[]'></div><br />");
    });
});
</script>
<div class="left marginT20"><input type="button" id="addRow" value="Add column" class="button red" /></div>
<span class="row_1 left marginT20"><input type="submit" name="submit" value="Add" class="button red" /></span>
</div>
</form>

Concatenating user data into an SQL string. That's a brilliant idea! — eggyal, Dec 27 '13 at 11:19
[Why you shouldn't use `mysql_*` functions anymore](http://stackoverflow.com/q/12859942/1612146) — George, Dec 27 '13 at 11:20
Put space between the keywords and column names while creating the sql — Nouphal.M, Dec 27 '13 at 11:23

score 0 · Answer 1 · answered Dec 27 '13 at 11:28

0

Check for your SQL syntax.

Change from

$Sql.=$_POST['column'][$i] .$_POST['data_type'][$i] ."(".$_POST['size'][$i].")";

to this

$Sql.=$_POST['column'][$i]." ".$_POST['data_type'][$i] ."(".$_POST['size'][$i].")";

answered Dec 27 '13 at 11:28

Nouphal.M

6,304
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Thanks for the reply. it worked and the idea of adding one NOT NULL is also better idea. Is there any better practice for te solution of such task.Please suggest. – Code4Code Dec 27 '13 at 13:28

creating table in database dynamically

1 Answers1