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我試過了所有的可能性。我無法讀取php.Am中的json對象,我正確地將json數據傳遞給PHP?我已經使用$ _POST,$ _ REQUEST和json_decode來獲取data.Nothing作品。如何獲取PHP中的json數據? Controller.js如何將json數據從角度js傳遞到php
if (userValid==true)
{
var data = {"firstName":user.firstName,
"lastName":user.lastName,
"mobileNumber":user.mobileNumber,
"email":user.email,
"type":user.type,
"password":user.password
};
$http.post("http://localhost/insertUser.php",data)
.then(function(response){
alert(response.data);
});
}
insertUser.php
<?php
header("Access-Control-Allow-Origin: *");
header("Content-Type: application/json; charset=UTF-8");
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn) {
die('Could not connect: ' . mysql_error());
}
$user = json_decode(file_get_contents("php://input"));
$firstName = mysql_real_escape_string($user->firstName);
$lastName = mysql_real_escape_string($user->lastName);
$mobileNumber = mysql_real_escape_string($user->mobileNumber);
$email = mysql_real_escape_string($user->email);
$type = mysql_real_escape_string($user->type);
$password = mysql_real_escape_string($user->password);
$sql = "INSERT INTO user ".
"VALUES ('".$firstName."','".$lastName."','".$mobileNumber."','".$email."','".$type."','".$password."')";
mysql_select_db('user_details');
$retval = mysql_query($sql, $conn);
if(! $retval) {
die('Could not enter data: ' . mysql_error());
}
echo "Entered data successfully\n";
mysql_close($conn);
?>
你得到的任何錯誤? –
@ujjwal:沒有。我是否正確地在js $ http服務中傳遞json並在php中正確讀取它? – Vimal