如果没有从下拉列表中选择值，则允许textbox使用php、mysql、javascript添加到mysql表。
如果没有从下拉列表中选择值，则允许textbox使用php、mysql、javascript添加到mysql表。
提问于 2012-08-31 05:44:04
嗨，我正在为国家，州，城市使用ajax三重下拉列表，参考链接是：http://roshanbh.com.np/2008/01/populate-triple-drop-down-list-change-options-value-from-database-using-ajax-and-php.html。它成功地工作了，但是我需要如果一个州在db表中没有城市，那么会出现一个新的文本框，输入的值存储在php mysql中。我实现的是什么编码。请给我一些想法。
代码：
Ajax：
<script language="javascript" type="text/javascript">
function getXMLHTTP() {
var xmlhttp = false;
try {
xmlhttp = new XMLHttpRequest();
} catch (e) {
try {
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
} catch (e) {
try {
xmlhttp = new ActiveXObject("Msxml2.XMLHTTP");
} catch (e1) {
xmlhttp = false;
}
}
}
return xmlhttp;
}
function getState(countryId) {
var strURL = "findState.php?country=" + countryId;
var req = getXMLHTTP();
if (req) {
req.onreadystatechange = function () {
if (req.readyState == 4) {
// only if "OK"
if (req.status == 200) {
document.getElementById('statediv').innerHTML = req.responseText;
} else {
alert("There was a problem while using XMLHTTP:\n" + req.statusText);
}
}
}
req.open("GET", strURL, true);
req.send(null);
}
}
function getCity(countryId, stateId) {
var strURL = "findCity.php?country=" + countryId + "&state=" + stateId;
var req = getXMLHTTP();
if (req) {
req.onreadystatechange = function () {
if (req.readyState == 4) {
// only if "OK"
if (req.status == 200) {
document.getElementById('citydiv').innerHTML = req.responseText;
} else {
alert("There was a problem while using XMLHTTP:\n" + req.statusText);
}
}
}
req.open("GET", strURL, true);
req.send(null);
}
}
</script>表格：
<form method="post" action="" name="form1">
<table width="60%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td width="150">Country</td>
<td width="150"><select name="country" onChange="getState(this.value)">
<option value="">Select Country</option>
<option value="1">USA</option>
<option value="2">Canada</option>
</select></td>
</tr>
<tr style="">
<td>State</td>
<td ><div id="statediv"><select name="state" >
<option>Select Country First</option>
</select></div></td>
</tr>
<tr style="">
<td>City</td>
<td ><div id="citydiv"><select name="city">
<option>Select State First</option>
</select></div></td>
</tr>
<tr>
<td> </td>
<td> </td>
</tr>
<tr>
<td> </td>
<td> </td>
</tr>
</table>
</form>findstate.php
<?php
include('config.php');
$country = intval($_GET['country']);
$query = "SELECT id,statename FROM state WHERE countryid='$country'";
$result = mysql_query($query);
?>
<select name="state" onchange="getCity(<?php echo $country?>,this.value)">
<option>Select State</option>
<?php while($row=mysql_fetch_array($result)) { ?>
<option value=<?php echo $row['id']?>><?php echo $row['statename']?></option>
<?php } ?>
</select>findcity.php
<?php
include('config.php');
$countryId = intval($_GET['country']);
$stateId = intval($_GET['state']);
$query = "SELECT id,city FROM city WHERE stateid='$stateId'";
$result = mysql_query($query);
?>
<select name="city">
<option>Select City</option>
<?php while($row=mysql_fetch_array($result)) { ?>
<option value><?php echo $row['city']?></option>
<?php } ?>
</select>我想要显示，如果在任何州没有城市，那么一个文本框将出现，它的值存储到db。
回答 2
Stack Overflow用户
发布于 2012-08-31 05:49:49
把这个放在findcity.php文件中
<?php
include('config.php');
$countryId=intval($_GET['country']);
$stateId=intval($_GET['state']);
$query="SELECT id,city FROM city WHERE stateid='$stateId'";
$result=mysql_query($query);
if(mysql_num_rows($result) == 0) // no cities found
{
echo '<input type="textbox" name="city" />';
}
else // show select box
{
?>
<select name="city">
<option>Select City</option>
<?php while($row=mysql_fetch_array($result)) { ?>
<option value><?php echo $row['city']?></option>
<?php } ?>
</select>
<?php
}
?>Stack Overflow用户
发布于 2012-08-31 05:53:27
你的findstate.php变成了
<?php
include('config.php');
$country=intval($_GET['country']);
$query="SELECT id,statename FROM state WHERE countryid='$country'";
$result=mysqli_query($query);
if(mysqli_num_rows($result)==0):?>
<input type="text" name="state" value="" />
<? endif; ?>
<select name="state" onchange="getCity(<?=$country ?>,this.value)">
<option>Select State</option>
<?php while($row=mysqli_fetch_array($result)): ?>
<option value=<?= $row['id']?>><?= $row['statename']?></option>
<?php endwhile; ?>
</select>同样，也要更改您的findcity.php。
建议
不再使用mysql*，使用mysqli*或PDO，因为mysql*被破坏了。链接
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/12209524
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