phpmady Posted August 2, 2011 Share Posted August 2, 2011 Hi, I am calling the ajax twice in a page, <div class="login_form"> <div class="row"> <span class="label">رقم الجوال<span class="men">*</span></span> <span class="formw"> <input type="text" class="textbox_forms required" name="SMS_Mobile" id="SMS_Mobile" size="50" onblur="[b]chkUser[/b](this.value);" dir="ltr"></span> </div> <div class="row" id="[b]user_already[/b]"> </div> <div class="row" id="choose_any_type"> <span class="label">Send By</span> <span class="formw"> <span> م : <input type="radio" value="format_message" name="message_type" onclick="call_format_message();"> </span> <span> ش : <input type="radio" value="custom_message" name="message_type" onclick="call_custom_message();"> </span> </span> </div> <div id="div_format_message" style="display:none;"> <div class="row1"><span class="label">Format Message</span> <span class="formw" dir=""> <select name="F_ID" id="gt" onchange="[b]ajax_call_format_message[/b](this.value);"> <option value="">Select SMS Template</option> <!-- START BLOCK : format_msg_send_one --> <option value="{F_ID}" {for_selected}>{F_Title}</option> <!-- END BLOCK : format_msg_send_one --> </select> </span> </div> <div class="row" id="show_div_format_message"> </div> </div> </div> when i am calling the chkUser actually am asking to bring the ajax content in the div called <div class="row" id="user_already"> </div> but the content is placing the div called <div class="row" id="show_div_format_message"> </div> Why its placing the content in a different location. Here is my js for chkUser // JavaScript Document var xmlhttp; //function namechange(name_fieldvalue) function chkUser(mobile_number) { xmlhttp=GetXmlHttpObject(); if (xmlhttp==null) { alert ("Your browser does not support XMLHTTP!"); return; } var url="ajax_user_operate.php"; url=url+"?mobile_number="+mobile_number; xmlhttp.onreadystatechange=stateChanged; xmlhttp.open("GET",url,true); xmlhttp.send(null); } function stateChanged() { if (xmlhttp.readyState==4) { document.getElementById("user_already").innerHTML=xmlhttp.responseText; } } function GetXmlHttpObject() { if (window.XMLHttpRequest) { // code for IE7+, Firefox, Chrome, Opera, Safari return new XMLHttpRequest(); } if (window.ActiveXObject) { // code for IE6, IE5 return new ActiveXObject("Microsoft.XMLHTTP"); } return null; } and js for ajax_call_format_message(); // JavaScript Document var xmlhttp; //function namechange(name_fieldvalue) function ajax_call_format_message(F_ID) { xmlhttp=GetXmlHttpObject(); if (xmlhttp==null) { alert ("Your browser does not support XMLHTTP!"); return; } var url="ajax_show_format_message.php"; url=url+"?F_ID="+F_ID; xmlhttp.onreadystatechange=stateChanged; xmlhttp.open("GET",url,true); xmlhttp.send(null); } function stateChanged() { if (xmlhttp.readyState==4) { document.getElementById("show_div_format_message").innerHTML=xmlhttp.responseText; } } function GetXmlHttpObject() { if (window.XMLHttpRequest) { // code for IE7+, Firefox, Chrome, Opera, Safari return new XMLHttpRequest(); } if (window.ActiveXObject) { // code for IE6, IE5 return new ActiveXObject("Microsoft.XMLHTTP"); } return null; } I seriously doubt the Js file, am i doing wrong.. Thanks, Quote Link to comment Share on other sites More sharing options...
sunfighter Posted August 2, 2011 Share Posted August 2, 2011 This line places the ajax response into the div: document.getElementById("show_div_format_message").innerHTML=xmlhttp.responseText; And yes this places it in <div id="show_div_format_message"> Change that line to document.getElementById("user_already").innerHTML=xmlhttp.responseText; to put it where you wanted. Quote Link to comment Share on other sites More sharing options...
phpmady Posted August 2, 2011 Author Share Posted August 2, 2011 Hi, As u said, i have placed my code ..if u go through the js code i have made, u can find that.. Thanks Quote Link to comment Share on other sites More sharing options...
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