我正在尝试使用ajax将表单数据存储到数据库中,但它没有显示任何成功也没有任何错误。 这是我的代码。
<form method="POST" id="add_user" name='reg' >
<fieldset>
<legend>Student information:-</legend>
<ul>
<li>
<label> FirstName: </label><input type="text" id="name" name="name" required>
<span id='error' style="display:none;color:red;"> Only alphabets </span>
</li>
<li>
<label> LastName: </label><input type="text" id="lname" name="lname" required>
<span id='error1' style="display:none;color:red;"> Only alphabets </span>
</li>
<li>
<label>Username:</label>
<input type="text" id="username" name="username"/>
< /li>
<li>
<label>Password:</label>
<input type="password" id="password" name="password"/>
</li>
<label>
Gender: </label>
<input type="radio" id='gender' name="gender" value="male" required> Male
<input type="radio" name="gender" id='gender' value="female" required> Female
<input type="radio" name="gender" id='gender' value="other" required> Other
<li>
<label>
Email: </label>
<input id="email" type="text" name="email" required>
<span id='error2' style="display:none;color:red;"> Invalid email </span>
</li>
<li>
<label> Mobile:</label>
<input id="mobile" type="text" maxlength="10" name="mobile" required >
<span id='error3' style="display:none;color:red;"> only digits </span>
</li>
<li>
address: <textarea name="address" type="text" rows="3" cols="40"> </textarea>
</li>
</ul>
<p><button class = 'button' type="submit" id='submit'>Add User</button></p>
</fieldset>
</form>
此表单中我输入存储到数据库中的任何值。 这是我的js文件,它使用ajax函数发送数据inext文件,该文件将结果存储到数据库中 的 serve.js
$(document).ready(function(){
$(document).on('submit','#add_user',function(e){
var form_data = $('#add_user').serialize();
var request = $.ajax({
url: 'fun.php?job=add',
cache : false,
data : form_data,
dataType : 'json',
contentType : 'application/json; charset=utf-8',
type : 'get'
});
request.done(function(output){
if (output.result == 'success'){
var name = $('#fname').val();
show_message("User '" + name + "' added successfully.", 'success' );
}, true);
} else{
show_message('Add request failed','error');
};
});
});
fun.php
if ($job != ''){
// Connect to database
$db_connection = mysqli_connect($db_server, $db_username, $db_password, $db_name);
if (mysqli_connect_errno()){
$result = 'error';
$message = 'Failed to connect to database: ' . mysqli_connect_error();
$job = '';
}
if ($job == 'add'){
/ / Add user
$query = "INSERT INTO oops ";
if (isset($_GET['name'])) { $query .= "name = '" . mysqli_real_escape_string($db_connection, $_GET['name']) . "', "; }
if (isset($_GET['lname'])) { $query .= "lname = '" . mysqli_real_escape_string($db_connection, $_GET['lname']) . "', "; }
if (isset($_GET['username'])) { $query .= "username = '" . mysqli_real_escape_string($db_connection, $_GET['username']) . "', "; }
if (isset($_GET['password'])) { $query .= "password = '" . mysqli_real_escape_string($db_connection, $_GET['password']) . "', "; }
if (isset($_GET['gender'])) { $query .= "gender = '" . mysqli_real_escape_string($db_connection, $_GET['gender']) . "', "; }
if (isset($_GET['email'])) { $query .= "email = '" . mysqli_real_escape_string($db_connection, $_GET['email']) . "', "; }
if (isset($_GET['mobile'])) { $query .= "mobile = '" . mysqli_real_escape_string($db_connection, $_GET['mobile']) . "', "; }
if (isset($_GET['address'])) { $query .= "address = '" . mysqli_real_escape_string($db_connection, $_GET['address']) . "'"; }
$query = mysqli_query($db_connection, $query);
if (!$query){
$result = 'error';
$message = 'query error';
} else {
$result = 'success';
$message = 'query success';
}
}
// Close database connection
mysqli_close($db_connection);
}
// Prepare data
$data = array(
"result" => $result,
"message" => $message,
"data" => $mysql_data
);
// Convert PHP array to JSON array
$json_data = json_encode($data);
print $json_data;
?>
如果您在我的代码中发现任何错误,我会遗漏一些内容,请帮忙。
答案 0 :(得分:1)
因为您在表单中使用post方法:
<form method="POST" id="add_user" name='reg' >
并试图通过get:
接收参数isset($_GET['name'])
到处使用post方法
并且在jQuery中你需要设置:
type: "POST"