[英]Why is my dynamic dropdown list not working?
假定下拉列表是从另一个表的列中获取记录,但是当前没有记录出现。 另外,如果用户在列表中找不到想要的内容,我将需要在下拉列表中显示一个“其他”选项,以便用户键入。 这是我的代码:
<script type="text/javascript">
function showfield(name){
if(name=='Other')document.getElementById('div1').innerHTML='Please Specify: <input type="text" name="other" />';
else document.getElementById('div1').innerHTML='';
}
</script>
<label for="issue_type">Issue Type</label>
<?php
include ("../db/dbConn.php");
$sql = "SELECT issue_type FROM issue where deleted =0";
$result=mysql_query($sql);
echo '<select class ="form-control" type="text" name="issue_type" id="issue_type" onchange="showfield(this.options[this.selectedIndex].value)" >';
while ($row = mysql_fetch_array($result))
{
echo "<option value='".$row['issue_type']."'>".$row['issue_type']." </option>";
}
echo "</select>";
?>
<div id="div1"></div>
将html部分分配给变量,然后将其回显。 尝试
<script type="text/javascript">
function showfield(name){
if(name=='Other')document.getElementById('div1').innerHTML='Please Specify: <input type="text" name="other" />';
else document.getElementById('div1').innerHTML='';
}
</script>
<label for="issue_type">Issue Type</label>
<?php
include ("../db/dbConn.php");
$sql = "SELECT issue_type FROM issue where deleted =0";
$result=mysql_query($sql);
$htm = '';
$htm .='<select class ="form-control" type="text" name="issue_type" id="issue_type" onchange="showfield(this.options[this.selectedIndex].value)" >';
while ($row = mysql_fetch_array($result))
{
$htm .="<option value='".$row['issue_type']."'>".$row['issue_type']." </option>";
}
$htm .= "<option value='Other'>Other</option>"; // add other option
$htm .="</select>";
echo $htm;
?>
<div id="div1"></div>
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