Creating table with variable name php mysql

Question

I'am trying to create table with variable name and columns via forms here's my query

$execute =  mysql_query('CREATE TABLE '.$tName.'(
            '.$cName.' '.$cType.' )');

And it's creating table with name from variable $tName but it doesnt create the columns.

Are you repeating this same query for adding columns? Once the table is created you need to `ALTER` it to add columns. — Marshall Davis, Apr 14 '15 at 18:11
No, it's just a sample, i know that i can modify it, but i have to create table with theese columns. — Patryk Brejdak, Apr 14 '15 at 18:16
Turn on error reporting or check your logs and see if you are getting any errors returned. From reading this one line I don't see why it wouldn't work. — Marshall Davis, Apr 14 '15 at 18:22
@ToothlessRebel checked theese variables has their values and there is no any error, it just creating table without column — Patryk Brejdak, Apr 14 '15 at 18:27
What is the value of `$execute` after this? It should be `true` or `false` for this type of query. If `false` then there was a MySQL issue, if `true` I am well and truly stumped. — Marshall Davis, Apr 14 '15 at 18:32

score 0 · Answer 1 · answered Apr 14 '15 at 20:10

0

Try this:

$sql= "CREATE TABLE $tName($cName $cType(45),$cName $cType(50))";

//echo $sql;
$op =mysql_query($sql);

answered Apr 14 '15 at 20:10

akr

739
4
15

score 0 · Answer 2 · answered Sep 11 '16 at 08:15

0

OK try this, I think it gives the general idea

<?php
include 'connect-db.php';
$tbl_name='kerry';
$field1='name';
$field2='value';
$field3='date';
mysql_query("CREATE TABLE `{$tbl_name}`
(
$field1 varchar(15),
$field2 int,
$field3 date
)");
?>

answered Sep 11 '16 at 08:15

kerry

629
1
5
9

Out of interest why was this question downvoted - it seems clear and useful – kerry Sep 11 '16 at 08:22

Creating table with variable name php mysql

2 Answers2

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