little problem coming in this code.

Posted on 2009-06-27
Last Modified: 2012-05-07
there is a little problem coming in this code.
1) when i comment the form tags line 25 and 79 then 28 and 30

line code will not work and line 79 generate php code will

2) when i uncommented the form tags then 28 and 30 line code

will work and line 79 code will not work. the url becomes very

awkward. in 2nd case, on clicking the button of generate php

code, the page gets post and nothing happens and the url

becomes like  http://localhost/new/12_table_create/dropdown_values/new.php?mname=&tname=&fname=&dtype=varchar&ftype=textbox&foptions=NULL&flength=&b=Generate+php+code

this query string gets added , why??

this is the js function
function rt(){
var tname = document.getElementById("tname").value;
var fname = document.getElementById("fname").value;
var dtype = document.getElementById("dtype").value;
var ftype = document.getElementById("ftype").value;
var foptions = document.getElementById("foptions").value;
var flength = document.getElementById("flength").value;
var obj = document.getElementById("show3");'get','php_code_generator.php?tname='+tname+'&fname='+fname+'&dtype='+dtype+'&ftype='+ftype+'&foptions='+foptions+'&flength='+flength);
      if(h.readyState == 4){

$host = "localhost";

$username = "root";

$password = "";

$database = "cmsbuilder";

$con = mysql_connect($host,$username,$password) or die("could not connect".mysql_error());

$db = mysql_select_db($database,$con);


<script type="text/javascript" language="javascript" src="1.js"></script>

<script type="text/javascript">

function number(){

	var thelist=document.getElementById("theList");







<form name="frmcopytext">

	<table border="1" bgcolor="#CCCCCC">

	<tr><td colspan="3" bgcolor="#CCCCCC"><h2>Add Table or fields to the Database</h2></td></tr>

	<tr><td>Menu Name</td><td colspan="2"><input type="text" onKeyUp="this.form.tname.value = this.value; " name="mname" id="mname" /></td></tr>

	<tr><td colspan="3"><b>Write new table name here or Select from the dropdown</b></tr>

	<tr><td>Table Name</td><td><input type="text" onKeyUp="this.form.mname.value = this.value; " name="tname" id="tname" /></td><td>

	<select id="theList" onchange="number()">


<?php //here i have written the code for 1 table not to be displayed in the dropdown.

$tables_to_hide = array('field_types', 'login');

$res = mysql_query("show tables") or die(mysql_error());

while ($row = mysql_fetch_array($res)){


    if (in_array($row[0], $tables_to_hide)) continue;

   $my_tables[] = $row[0];


foreach ($my_tables as $tablename){

   echo "<option>$tablename</option>\n";   




	<tr><td>Field name</td><td colspan="2"><input type="text" name="fname" id="fname" /></td></tr>

	<tr><td>Data Type</td>

	<td colspan="2">

		<select name="dtype" id="dtype">





	<tr><td>Field Type</td>

	<td colspan="2">

		<select name="ftype" id="ftype">







	<tr><td>Field Options</td><td colspan="2"><select name="foptions" id="foptions">


			<option>NOT NULL</option>

			<option>NOT NULL DEFAULT 0</option>

			<option>NOT NULL DEFAULT ''</option>

			<option>NOT NULL DEFAULT '(undefined)'</option>


			<option>NOT NULL UNIQUE KEY</option>

			<option>NOT NULL KEY</option>


	<tr><td>Field Length</td><td colspan="2"><input type="text" name="flength" id="flength" /></td></tr>

<tr><td colspan="3"><input type="submit" name="a" value="Add to the database" onclick="javascript:createtable();" /><input type="submit" name="b" value="Generate php code" onclick="javascript:rt();" /></td></tr>



<div id="show2"></div>

<div id="show3"></div>

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Question by:designersx
1 Comment
LVL 17

Accepted Solution

nanharbison earned 250 total points
ID: 24727669
You should move this question to the javascript zone. Some of the lines you comment out involve javascript, not PHP.

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