這是我的角碼。形式提交代碼。當點擊提交按鈕。 JSON Parse error: Unrecognized token '<'
此錯誤將顯示。空記錄將保存在數據庫中。我也爲此添加了html代碼和PHP服務器端代碼。JSON解析錯誤:無法識別的令牌'<'處於角度
$scope.submitForm = function() {
$http({
method : 'POST',
url : 'http://localhost/youtubewebservice/checkOutt.php',
data : $scope.user,
dataType: 'json',
headers : {'Content-Type': 'application/x-www-form-urlencoded'}
})
.success(function(data) {
if (data.errors) {
$scope.errorinputFName = data.errors.inputFName;
$scope.errorinputLName = data.errors.inputLName;
}
});
};
HTML代碼
<form name="userForm" ng-submit="submitForm()">
<div class="form-group">
<label>Name</label>
<input type="text" name="inputFName" class="form-control" ng-model="user.inputFName">
<span ng-show="errorName">{{errorName}}</span>
</div>
<div class="form-group">
<label>Email</label>
<input type="text" name="inputLName" class="form-control" ng-model="user.inputLName">
<span ng-show="errorEmail">{{errorEmail}}</span>
</div>
<button type="submit" class="btn btn-primary">Submit</button>
<div id="sendmessageresponse"></div>
</form>
** PHP代碼**
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json;charset=UTF-8");
$data = json_decode(file_get_contents("php://input"));
$inputFName = mysql_real_escape_string($data->inputFName);
$inputLName = mysql_real_escape_string($data->inputLName);
$con = mysql_connect('localhost', 'root', '');
mysql_select_db('look4com_lk', $con);
$qry_em = 'select count(*) as cnt from checkout where chkID ="' . $chkID . '"';
$qry_res = mysql_query($qry_em);
$res = mysql_fetch_assoc($qry_res);
if ($res['cnt'] == 0) {
$qry = 'INSERT INTO checkout (inputFName,inputLName) values ("' . $inputFName . '","' . $inputLName . '")';
$qry_res = mysql_query($qry);
if ($qry_res) {
$arr = array('msg' => "User Created Successfully!!!", 'error' => '');
$jsn = json_encode($arr);
print_r($jsn);
} else {
$arr = array('msg' => "", 'error' => 'Error In inserting record');
$jsn = json_encode($arr);
print_r($jsn);
}
} else {
$arr = array('msg' => "", 'error' => 'User Already exists with same email');
$jsn = json_encode($arr);
print_r($jsn);
}
檢查什麼正在通過發送到服務器提琴手 – Sajeetharan
沒有問題。 – moni123
粘貼正在發送的json的輸出。 – Samar