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我花了大約2個小時來改變腳本並找到其他方法,但沒有運氣。如果有人能夠指出我的腳本中的任何問題,那就太棒了。非常感謝。SQL更新或插入查詢不起作用
我已經更新了腳本,希望這可以提供更好的理解和更多的細節。
<?php
session_start();
if(!isset($_SESSION["user"]) or !is_array($_SESSION["user"]) or empty($_SESSION["user"])) {
// redirect to login page
}
$dbhost = "localhost";
$dbname = "**";
$dbuser = "root";
$dbpass = "**";
// database connection
$conn = new PDO("mysql:host=$dbhost;dbname=$dbname",$dbuser,$dbpass);
// new data
error_reporting(E_ALL);
$work = $_POST['work'];
$education = $_POST['education'];
$location = $_POST['location'];
$relationship = $_POST['relationship'];
$phone = $_POST['phone'];
$email = $_POST['email'];
$website = $_POST['website'];
$language = $_POST['language'];
$skill = $_POST['skill'];
$tumblr = $_POST['tumblr'];
$instagram = $_POST['instagram'];
$skype = $_POST['skype'];
$facebook = $_POST['facebook'];
$youtube = $_POST['youtube'];
$twitter = $_POST['twitter'];
$about = $_POST['about'];
$user_id = $_SESSION['user']['id'];
echo $work."</br>";
echo $education."</br>";
echo $location."</br>";
echo $relationship."</br>";
echo $phone."</br>";
echo $email."</br>";
echo $website."</br>";
echo $language."</br>";
echo $skill."</br>";
echo $tumblr."</br>";
echo $instagram."</br>";
echo $skype."</br>";
echo $facebook."</br>";
echo $youtube."</br>";
echo $twitter."</br>";
echo $about."</br>";
echo $user_id;
/*
"IF EXISTS (SELECT * FROM Table1 WHERE Column1=’SomeValue’)
UPDATE Table1 SET (…) WHERE Column1=’SomeValue’
ELSE
INSERT INTO Table1 VALUES (…)"
*/
$sql = "IF EXISTS (SELECT * FROM user_details WHERE user_id=".$_SESSION['user']['id'].")
UPDATE user_details SET
(work='".$work."',education='".$education."',location='".$location."',relationship='".$relationship."',phone='".$phone."',email='".$email."',website='".$website."',language='".$language."',skill='".$skill."',tumblr='".$tumblr."',instagram='".$instagram."',skype='".$skype."',facebook='".$facebook."',youtube='".$youtube."',twitter='".$twitter."',about='".$about."')
WHERE user_id='".$user_id."'
ELSE INSERT INTO user_details (work,education,location,relationship,phone,email,website,language,skill,tumblr,instagram,skype,facebook,youtube,twitter,about,user_id)
VALUES
('".$work."','".$education."','".$location."','".$relationship."','".$phone."','".$email."','".$website."','".$language."','".$skill."','".$tumblr."','".$instagram."','".$skype."','".$facebook."','".$youtube."','".$twitter."','".$about."','".$user_id."')";
$q = $conn->prepare($sql);
$q->execute();
var_dump(mysql_error())
// header('Location: ../usr_settings.php');
?>
那麼這裏沒有錯誤信息。所有的領域正在填寫和張貼形式,我通過回聲方法檢查。沒有SQL錯誤或任何東西。 我想更新一個數據庫表,如果它們不是帶有用戶ID的查詢,我需要插入一個新的。 –
而且,當然,您正在使用mysql_query或類似的東西運行該查詢? –
是的,MySQL不是mysqli –