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, Link to comment https://forums.phpfreaks.com/topic/243625-calling-ajax-twice-in-a-page-causing-problem/ 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. Link to comment https://forums.phpfreaks.com/topic/243625-calling-ajax-twice-in-a-page-causing-problem/#findComment-1250897 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 Link to comment https://forums.phpfreaks.com/topic/243625-calling-ajax-twice-in-a-page-causing-problem/#findComment-1250920 Share on other sites More sharing options...
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