grs5211 Posted May 26, 2009 Share Posted May 26, 2009 I have filled an Input box with a value obtained by getting a cookie value at the start of code. This works fine. Now I want to input "aCode" with this value within the same form. Here is the code to get the cookie, which works, but the rest does not. NB. I am new to PHP. <head> function checkCookie() { if(document.getElementById("userid").value == "") { cookieCASUserIDValue = getCookie("CUserID"); if(cookieCASUserIDValue != null) { document.getElementById("userid").value = getCookie("CUserID"); } } </head> <body class="gradient" onLoad="checkCookie();"> <form id="gData" name="gData" method="POST"> <input type="text" id="userid" name="userid" size=8 >' /> <input type="text" name="aCode" id="aCode" style="display: none;" value=document.getElementById("userid").value> </input> </form> } At this stage I wish to get the value of the element "userid" and and display it. Here is my code that does not wotk. <?php error_reporting(1); // Only display error messages, not warnings and notices. echo "UserID = " . $_REQUEST["userid"]; . "<br>"; ?> Can anyone help! Link to comment https://forums.phpfreaks.com/topic/159723-get-the-value-of-an-input-variable-and-display-it-on-another-form-on-the-same-pa/ Share on other sites More sharing options...
Brian W Posted May 26, 2009 Share Posted May 26, 2009 if you use the code tags around your php or even html snippets, you will likely get more help [ code ]Like this but without the spaces in the tags[ /code ] would be Like this but without the spaces in the tags Try: <?php error_reporting(1); // Only display error messages, not warnings and notices. echo "UserID = " . $_REQUEST["CUserID"]; . "<br>"; ?> Else, you could do this in java script also like the below: <!--if you haven't set the cookie somewhere else, this isn't going to work--> <head> <script language="javascript" type="text/javascript"> function checkCookie() { if(document.getElementById("userid").value == ""){ cookieCASUserIDValue = getCookie("CUserID"); if(cookieCASUserIDValue != null){ document.getElementById("userid").value = getCookie("CUserID"); document.getElementById("aCode").value = getCookie("CUserID"); document.getElementById("userid_display").innerHTML = "User Id = " + getCookie("CUserID"); } } } function getCookie(c_name) { if (document.cookie.length>0) { c_start=document.cookie.indexOf(c_name + "="); if (c_start!=-1) { c_start=c_start + c_name.length+1; c_end=document.cookie.indexOf(";",c_start); if (c_end==-1) c_end=document.cookie.length; return unescape(document.cookie.substring(c_start,c_end)); } } return ""; } </script> </head> <body class="gradient" onLoad="checkCookie();"> <form id="gData" name="gData" method="POST"> <input type="text" id="userid" name="userid" size=8 > <input type="hidden" name="aCode" id="aCode" value=""> </form> <div id="userid_display"></div> </body> In that example, I've cleaned up your code, fixed things A LOT . Good luck and I wish you the best in learning. Link to comment https://forums.phpfreaks.com/topic/159723-get-the-value-of-an-input-variable-and-display-it-on-another-form-on-the-same-pa/#findComment-842503 Share on other sites More sharing options...
grs5211 Posted May 26, 2009 Author Share Posted May 26, 2009 Thanks, that solved the problem. Link to comment https://forums.phpfreaks.com/topic/159723-get-the-value-of-an-input-variable-and-display-it-on-another-form-on-the-same-pa/#findComment-842537 Share on other sites More sharing options...
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