第1页：index.php
脚本ajax
<script>
function fcall(clusters) {
var clusters = document.getElementById("clusters").value;
var request = $.ajax({
type: "get",
url: "chart.php",
data: {clusters: clusters}
});
request.done( function( msg ) {
$("#page").html(msg);
});
}
</script>
脚本php
$sql = mysql_query("SELECT * FROM tree_hie GROUP BY no_cluster asc") or die(mysql_error());
while ($data1=mysql_fetch_array($sql)) {
$clusters = $data1['no_cluster'];
$sql2 = mysql_query("SELECT * FROM tree_hie where no_cluster = '$clusters'") or die(mysql_error());
while ($data2 = mysql_fetch_array($sql2)) {
<input type="button" id="clusters" name="clusters" onclick="fcall(<?php $data1['no_cluster']; ?>);" value="<?php echo $clusters; ?>">
}} /* for while
<div id="page">
</div>
第2页图表.php
$clusters = $_GET['clusters']."tez<br>";
echo $clusters."tez";
我有那个代码。。
我的问题。。
输出为链接，例如
集群1
集群2。。。。。etc
if i click button 'cluster 1', should on div page display 1
if i click button 'cluster 2' should on div page display 2 ...
BUT ..
if i click button 'cluster 1', on div page display 1
if i click button 'cluster 2' on div page display 1 too ,
still can't get value ..
你知道怎么修吗？
您有两个id相同的按钮。在函数调用中传递事件对象
<input type="button" id="clusters" name="clusters" onclick="fcall(event);" value="<?php echo $clusters; ?>">
并使用
function fcall(event) {
var clusters = event.target.value;
....
}