如果用户单击"全选"按钮,我只想检查页面上的所有复选框.
在身体里:
<button id="checkall">Select All</button>
jQuery的:
<head>
<script type="text/javascript" src="js/jquery.js"></script>
<script type="text/javascript">
$(document).ready(function(){
$('#checkall').click(function () {
$("input[type=checkbox]").each( function() {
$(this).attr('checked', true);
});
});
</script>
</head>
而且复选框是由一点点php生成的:
echo '<li>';
echo '<input type="checkbox" name="checkbox[]" value="'.$x.'" />'.$x.'<br/>';
echo '</li>';
谁能告诉我我做错了什么？
我有四个变量,我想通过AJAX传递给由同一个页面上一些PHP进行处理:newJudgeName,newJudgeSection,newJudgeStatus和originalJudgeName.成功函数回显它们并且它们是正确的值,它只是我的PHP没有获取newJudgeStatus变量.我在AJAX请求数据线切换newJudgeStatus与newJudgeName然后值送到就好了(我可以看到它在法官下名称DB)......只有当它是在原来的点在Ajax请求它不起作用.我是Ajax的新手.任何帮助将非常感激.
AJAX:
$.ajax({
type: "POST",
url: "test.php",
data: 'newJudgeName=' + newJudgeName + '&newJudgeSection=' + newJudgeSection + '&newJudgeStatus =' + newJudgeStatus + '&originalJudgeName=' + originalJudgeName,
success: function(){
alert('newJudgeName=' + newJudgeName + '&newJudgeSection=' + newJudgeSection + '&newJudgeStatus =' + newJudgeStatus +'&originalJudgeName=' + originalJudgeName);
}
});
PHP:
if(isset($_POST['newJudgeName'])){
$newJudgeName = $_POST['newJudgeName'];
$newJudgeSection = $_POST['newJudgeSection'];
$newJudgeStatus = $_POST['newJudgeStatus'];
$originalJudgeName = $_POST['originalJudgeName'];
$judgeID = judgeNametoID($originalJudgeName);
$con = mysql_connect("-","-","-");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
else {
// connected to database successfully …