我有一个有选项的表单,用户需要选择一个状态。如何用纯PHP验证服务器端这个?如果选择“选择状态”选项,则服务器将返回“无效”。
谢谢!
我的HTML代码(摘要):
<select id="stateSel" name="state" onclick="stateValidation()">
<option value="0">Select State</option> /*will return as invalid option*/
<option value="AL">Alabama</option>
<option value="AK">Alaska</option>
<option value="AZ">Arizona</option>
/*[etc, options for each state]*/
<option value="WY">Wyoming</option>
</select>
如果用户没有更改状态,则状态的默认值在加载时始终为0,然后您需要验证如下代码
以下代码是表格提交
<?php
if( $_POST['state'] == "0" ) {
echo "Please select the state field";
}else{
echo "Your selected state is ".$_POST['state'];//just for example purpose
}
die;
?>
要么
以下代码适用于JS功能
<select id="stateSel" name="state" onclick="stateValidation(this.id)">
<option value="0">Select State</option> /*will return as invalid option*/
<option value="AL">Alabama</option>
<option value="AK">Alaska</option>
<option value="AZ">Arizona</option>
<option value="WY">Wyoming</option>
</select>
function stateValidation(id){
var state = $('#'+id+'option:selected').val();
if(state == '0'){
alert("Please select the state field");
}else{
alert("Your selected state is "+state);
}
}