您好想问一下使用jquery传递参数的正确方法是什么我的参数来自html输入,下面是我的代码。
function ajax_post(){
var param = document.getElementById('def').value;
$('#chatlogs').load('main.php?id='+param);
}
我可以加载main.php,但无法回显param,这是我的main.php代码。
<?php echo $_GET['id'];?>
添加更多细节..我从HTML选择输入中获取参数然后在获取值之后,将main.php加载到id为#chatlogs的div中。
<select name="topic" id="def" class="form-control" style="width:450px;" onChange='ajax_post()'>
<option value="" selected>---</option>
<option value="Solder Short">Solder Short</option>
<option value="Insufficient Solder">Insufficient Solder</option>
<option value="Misaligned Component">Misaligned Component</option>
<option value="Missing Component">Missing Component</option>
<option value="Inverted Component">Inverted Component</option>
<option value="Pad Contamination">Pad Contamination</option>
</select>
<div id="chatlogs">
Please select a defect category...
</div>
答案 0 :(得分:0)
试试这个:
function ajax_post(){
var param = document.getElementById('def').value;
$('#chatlogs').load('main.php?id='+param);
}
答案 1 :(得分:0)
试试这个:
<select name="topic" id="def" class="form-control" style="width:450px;">
<option value="" selected>---</option>
<option value="Solder Short">Solder Short</option>
<option value="Insufficient Solder">Insufficient Solder</option>
<option value="Misaligned Component">Misaligned Component</option>
<option value="Missing Component">Missing Component</option>
<option value="Inverted Component">Inverted Component</option>
<option value="Pad Contamination">Pad Contamination</option>
</select>
<div id="chatlogs">
Please select a defect category...
</div>
<script>
$(document).ready(function(){
$('#def').on('change', function() {
var param = document.getElementById('def').value;
//alert(param);
$('#chatlogs').load('main.php?id=', param);
})
});
</script>