sql查詢未運行

-7

<?php 
include"connect.php"; 

    $name =  $_REQUEST['name']; 
    $address = $_REQUEST['address']; 
    $password =  $_REQUEST['password']; 
    $email = $_REQUEST['email']; 
    $number = $_REQUEST['number']; 
    $hear =  $_REQUEST['hear']; 
    $money = $_REQUEST['money']; 
    $artist = $_REQUEST['artist']; 
    $security = $_REQUEST['security']; 
    $city =  $_REQUEST['city']; 
    $pro = $_REQUEST['pro']; 
    $role =  $_REQUEST['role']; 
    $accname = $_REQUEST['accname']; 
    $accpass = $_REQUEST['accpass']; 
    $accno = $_REQUEST['accno']; 
    $seaccname = $_REQUEST['seaccname']; 
    $seaccpass = $_REQUEST['seaccpass']; 
    $seaccno = $_REQUEST['seaccno']; 
    $contracts = $_REQUEST['contracts']; 
    $statements = $_REQUEST['statements']; 

    $result = mysqli_query($con, "insert into signup set name='$name',email='$email',password='$password' ,address='$address',cell_phone_number='$number',heard_from='$hear',money_time='$money',fav_artist='$artist',fav_city='$city',security_question='$security',pro='$pro',pro_acc_name='$accname',pro_acc_password='$accpass',pro_acc_number='$accno',se_acc_name='$seaccname',se_acc_pass='$seaccpass',se_ac_number='$seaccno',rec_contract_copy='$contracts',rec_label_copy='$statements',role='$role'"); 

//$result= mysqli_query($con, "INSERT INTO signup (name, email,password, address,cell_phone_number, heard_from,money_time, fav_artist,fav_city,security_question, pro,pro_acc_name, pro_acc_password,pro_acc_number, se_acc_name,se_acc_pass, se_ac_number,rec_contract_copy, rec_label_copy,role) VALUES ('$name', '$email', '$password', '$address', '$number', '$hear', '$money', '$artist', '$security', '$city', '$pro', '$role', '$accname', '$accpass', '$accno', '$seaccname', '$seaccpass', '$seaccno', '$contracts', '$statements')"); 



    if($result==true) 
    { 
     //echo "<script>alert('user successfully added')</script>"; 
     echo "Success"; 

    } 
else{ 

    echo "Failed" ; 

} 


?>

來源

2017-10-04 Yash Singhal

然後檢查真正的錯誤。如果你不知道如何，請詢問。連接是未知的，如果所有的$ _REQUEST都有價值。 –

mysqli_error（$ con）的輸出是什麼？ – LordNeo

您的代碼易受[** SQL注入**]（https://en.wikipedia.org/wiki/SQL_injection）攻擊。您應該通過[** mysqli **]（https://secure.php.net/manual/en/mysqli.prepare.php）或[** PDO **]（https ：//secure.php.net/manual/en/pdo.prepared-statements.php）驅動程序。 [**這篇文章**]（https://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php）有一些很好的例子。 –

它不會運行，因爲您的查詢錯誤，您正在使用更新格式的insert語句。

$result= mysqli_query($con, "INSERT INTO signup (name, email,password, address,cell_phone_number, heard_from,money_time, fav_artist,fav_city,security_question, pro,pro_acc_name, pro_acc_password,pro_acc_number, se_acc_name,se_acc_pass, se_ac_number,rec_contract_copy, rec_label_copy,role) VALUES ('$name', '$email', '$password', '$address', '$number', '$hear', '$money', '$artist', '$security', '$city', '$pro', '$role', '$accname', '$accpass', '$accno', '$seaccname', '$seaccpass', '$seaccno', '$contracts', '$statements')");

進一步信息，開始使用預處理語句PDO

來源

2017-10-04 18:35:57

它看起來OK。問題可能出現在connect.php中，或$ _REQUEST中的名稱與發送請求的文件不同。同時檢查您要插入的數據庫中的屬性名稱是否與查詢中的名稱匹配。

來源

2017-10-04 18:40:25 Ana

sql查詢未運行

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