[英]How to get data from select tag using AJAX before submitting into PHP?
我想获取选择标记id =“ departmentselect”的数据，以便在提交表单之前将其值存储在mysql数据库中。 听说您将在提交表单之前使用AJAX获取数据。 因为当我选择“学院选择”标签及其对应的“部门选择”标签值时，它只会在数据库的部门中存储一个数字。
示例MYSQL数据库：
在mysql查询中，部门未获取JSON文件的值。 它只显示数字。
这是我的PHPCode
<!DOCTYPE html>
<html>
<head>
<title>Sample</title>
<link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
</head>
<body>
<form action="SignupProcess.php" method="POST" onsubmit="return check_password();" id="signupform">
<select id="collegeselect" name="collegeselect">
<option value="">College</option>
<option value="College of CAS">College of CAS</option>
</select>
<select id="departmentselect" name="departmentselect">
<option value="">Department</option>
</select>
</form>
</body>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<script src="https://code.jquery.com/jquery-2.1.1.min.js" type="text/javascript"></script>
<script src="GetCollegeJsonData.js"></script>
<script>
$('#signupform').submit(function(e))
{
if($('#departmentselect').val() != '')
{
$ajax
({
type: 'POST',
url: 'Signup.php?select' += select,
success: function(data)
{
alert(data);
}
});
}
else
{
alert('error');
}
e.preventDefault();
});
</script>
</html>
脚本类型这是使用ajax的脚本，但似乎没有任何作用。
<script>
$('#signupform').submit(function(e))
{
if($('#departmentselect').val() != '')
{
$ajax
({
type: 'POST',
url: 'Signup.php?select' += select,
success: function(data)
{
alert(data);
}
});
}
else
{
alert('error');
}
e.preventDefault();
});
</script>
JQUERY代码文件名GetCollegeJsonData.js
我从文件中获取JSON数据，并将其读入我的Jquery文件，然后使用脚本将该文件链接到我的php代码
//Get json
$('body').on('change', '#collegeselect', function() {
var selected_college = $(this).val();
$('#departmentselect').html('');
$.getJSON("CollegeAndDepartment.json", function(data) {
$.each(data[selected_college], function(key, val) {
$('#departmentselect').append("<option value='" + key + "'>" + val + "</option>");
});
});
})
其JSON文件
{
"College of CAS": ["Biology", "English", "LIACOM", "Library & Information Science", "Mass Communication", "Philosophy", "Political Science", "Psychology"]
}
我的Ajax功能不正确吗？
您的ajax方法代码具有语法错误，并且当您使用post方法时，您必须使用data选项（而非URL）来发布数据。 数据应位于json对象或查询字符串中。 你的ajax函数应该是
<script>
$('#signupform').submit(function(e))
{
if($('#departmentselect').val() != '')
{
$.ajax({
type: 'POST',
url: 'Signup.php',
data: {select: $('#departmentselect').val()},
success: function(data)
{
alert(data);
}
});
}
else
{
alert('error');
}
e.preventDefault();
});
</script>
我已经如下编辑您的代码。
Kindly give it a try.
<!DOCTYPE html>
<html>
<head>
<title>Sample</title>
<link rel="stylesheet" href="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
</head>
<body>
<form action="SignupProcess.php" method="POST" onsubmit="return check_password();" id="signupform">
<select id="collegeselect" name="collegeselect">
<option value="">College</option>
<option value="College of CAS">College of CAS</option>
</select>
<select id="departmentselect" name="departmentselect">
<option value="">Department</option>
</select>
</form>
</body>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.4/jquery.min.js"></script>
<script src="http://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
<script src="https://code.jquery.com/jquery-2.1.1.min.js" type="text/javascript"></script>
<script src="GetCollegeJsonData.js"></script>
<script>
$('#signupform').submit(function(e)
{
if($('#departmentselect').val() != '')
{
$.ajax
({
type: 'POST',
data:{select:$('#departmentselect').val()},
url: 'Signup.php',
success: function(data)
{
alert(data);
}
});
}
else
{
alert('error');
}
e.preventDefault();
});
</script>
</html>
For GetCollegeJsonData.js, ihave modified as follows:
//Get json
$('#collegeselect').on('change', function() {
var selected_college = $(this).val();
$('#departmentselect').html('');
$.getJSON("CollegeAndDepartment.json", function(data) {
$.each(data[selected_college], function(key, val) {
$('#departmentselect').append("<option value='" + val + "'>" + val + "</option>");
});
});
})
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