我有一个div通过ajax请求填充。 在div内是一种完成时应该使用相同类型的ajax请求来填充更多的div的形式。我用同样的方法来创建既但第二JavaScript不运行:阿贾克斯没有射击由阿贾克斯人口填充
第一个(工作):
<div class="content_text" id="searchbysurname">
<p><form name="searchbysurname">
<b>Search by Surname: </b><input class="inline" type="text" name="q">
<input type="submit"></form>
<script>
$('#searchbysurname form').submit(function(){
var data=$(this).serialize();
// post data
$.post('searchbysurname_test.php', data , function(returnData){
$('#resultstable').html(returnData)
})
return false; // stops browser from doing default submit process
});
</script>
<div id="resultstable"></div>
第二个(这是在resultstable DIV),这并不工作:
<? require_once("dbcontroller.php");
$db_handle = new DBController();
$q = ($_POST['q']);
$employees=array();
$sql = "SELECT employees.employeeid, employees.firstname, employees.surname FROM employees where UCASE(employees.surname) LIKE UCASE('%".$q."%')";
$employees = $db_handle->runQuery($sql); ?>
<table class="invisible">
<?
if(isset($employees) && !empty($employees)){
foreach($employees as $k=>$v) {
?>
<tr><td><?php echo $employees[$k]["firstname"]; ?> <?php echo $employees[$k]["surname"]; ?> </td>
<td><div id="viewemployeedetails<? echo $employees[$k]["employeeid"]?>">
<form>
<input type="hidden" name="id" value="<? echo $employees[$k]["employeeid"]?>">
<input type="submit" value="View">
</form>
</div></td>
<div id="mainpart"><b></b></div>
<script>
$('viewemployeedetails<? echo $employees[$k]["employeeid"]?> form').submit(function(){
var data=$(this).serialize();
// post data
$.post('viewemployeedetails.php', data , function(returnData){
$('#mainpart').html(returnData)
})
return false; // stops browser from doing default submit process
});
</script>
你缺少一个**#**在你的选择。 –