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mrhussein

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About mrhussein

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  1. thank u i found it and this is the correction : <?php while($data=mysql_fetch_array($query)){ ?><option value="<?php echo $data['id'] ?>"><?php echo $data['name'] ?></option><?php } ?>
  2. actully i changed the extntion to .php and did the following but the page dont want to apear: <html><head><script type="text/javascript">function showUser(str){if (str=="") { document.getElementById("txtHint").innerHTML=""; return; }if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Safari xmlhttp=new XMLHttpRequest(); }else {// code for IE6, IE5 xmlhttp=new ActiveXObject("Microsoft.XMLHTTP"); }xmlhttp.onreadystatechange=function() { if (xmlhttp.readyState==4 && xmlhttp.status==200) { document.getElementById("txtHint").innerHTML=
  3. Thank You Very Very Much your helpfull
  4. Hello, regarding to this example in your website: this example display data from mysql database\table with parameter (dropdown list),and i do input the choices to this list,what if i want same this but the choices in the (dropdown list) come from database\table as well ? anyone can help me plz? this example containing two files as the following: 1. HTML File: <html><head><script type="text/javascript">function showUser(str){if (str=="") { document.getElementById("txtHint").innerHTML=""; return; }if (window.XMLHttpRequest) {// code for IE7+, Firefox, Chrome, Opera, Sa
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