做了一个按条件筛选显示数据库列表的PHP页面,选择下拉框中条件后,现在点按钮后能进行筛选显示,但是刷新页面或再从其它页面转入进来时,筛选就失效了,如何才能将筛选显示保持不变(就是在刷新或其它页面转入进来时保持筛选条件),直到选下拉框首行点“筛选”按钮后再取消筛选条件。
这是我现在的代码
以下是刷新当前页面的数据库连接代码,$condition和$condition1是两个条件筛选变量。
$condition = " 1=1 ";//注意,前面有个空格
if(isset($_GET['shaixuan'])){
if($_GET['shaixuan']!='3'){
$condition.=" AND 标识='".$_GET['shaixuan']."'";
}}
$condition1 = " 1=1 ";//注意,前面有个空格
if(isset($_GET['shaixuan1'])){
if($_GET['shaixuan1']!='3'){
$condition1.=" AND 诊断意见2='".$_GET['shaixuan1'] . "'";
}}
$query_rsdb = "SELECT * FROM dengji where 检查日期='$today' AND $condition AND $condition1";
以下是两个下拉框筛选框和按钮的代码:
<form id="filter-form" action="listc.php" method="GET" enctype="multipart/form-data" name="form1">
<select name="shaixuan" style="width:50%;height:20px;font-size: 15px" id="shaixuan">
<option value="3">-报告状态-</option>
<option value="0" >已提交</option>
<option value="2" >未报告</option>
</select>
<select name="shaixuan1" style="width:50%;height:20px;font-size: 15px" id="shaixuan1" >
<option value=3>-医疗机构-</option>
<?php
$sql= "select name from hospital";//sql语句
$result = mysql_query($sql, $db_conn);//执行sql语句
while($row = mysql_fetch_array($result)) {
$name = $row['name'];
$template = $row['name'];
echo "<option value='$template' data-name='$name'>$template</option>";
}?>
</select>
</form>
以下是javascript代码:
<script language="javascript">
$('#filter-form').on('submit',function(event){
event.preventDefault();
var url = "http://127.0.0.1/listc.php?";
var shaixuan = 'shaixuan='+$('#shaixuan').val();
var shaixuan1 = 'shaixuan1='+$('#shaixuan1').val();
url += shaixuan + '&' + shaixuan1;
window.location.href = url;
});
</script>