问题
I am using jquery datatables to display results on a table and then passing ID as URL parameter, we able to display details of a particular record to user.
So far, this works fine.
However, I have been asked to pass a second url parameter called user and this is presenting problems.
When I run the code, it is no longer displaying details page.
Upon reviewing the code using firebug, I found an error that says:
john is undefined.
john.doe is the value being displayed on the url that looks more like this:
http://mylink.com/viewDetails.php?ID=1235&userName=john.doe
Any ideas what I am doing wrong?
return "<a target='tab' data-tab-name='#RequestDetails' data-tab-index='2' href='viewDetails.php?rID="
+ oObj.aData["ID"] + "&userName=" + <?php echo $user; ?> + "'> " + oObj.aData["ID"] + " </a>";
回答1:
Your code looks like this after being parsed by PHP:
<a target='tab' data-tab-name='#RequestDetails' data-tab-index='2' href='viewDetails.php?rID=" + oObj.aData["ID"] + "&userName=" + john + "'> " + oObj.aData["ID"] + " </a>
That means it's looking for a variable called john, which as you can see is undefined.
It should be this:
<a target='tab' data-tab-name='#RequestDetails' data-tab-index='2' href='viewDetails.php?rID=" + oObj.aData["ID"] + "&userName=<?php echo $user?>'> " + oObj.aData["ID"] + " </a>
回答2:
You need to but quotes around the php element as it is just going to write out the value of itself to the page. This way it becomes a string.
return "<a target='tab' data-tab-name='#RequestDetails' data-tab-index='2' href='viewDetails.php?rID="
+ oObj.aData["ID"] + "&userName=" + "<?php echo $user; ?>" + "'> " + oObj.aData["ID"] + " </a>";
Better put is this way...
return "<a target='tab' data-tab-name='#RequestDetails' data-tab-index='2' href='viewDetails.php?rID="
+ oObj.aData["ID"] + "&userName=<?php echo $user; ?>'> " + oObj.aData["ID"] + " </a>";
来源：https://stackoverflow.com/questions/24087258/value-is-coming-back-as-undefined