在向mysql插入值時獲取<br />

當使用php將值插入到MySQL時，我得到一條<br />語句。我曾嘗試更改插入語句以及回聲語句，但沒有任何工作。在向mysql插入值時獲取<br />

<?php 
    if($_SERVER['REQUEST_METHOD']=='POST'){ 
    $userfullname=$_POST['userfullname']; 
    $usernickname=$_POST['usernickname']; 
    $usermail=$_POST['usermail']; 
    $userusname=$_POST['userusname']; 
    $useruspassword=$_POST['useruspassword']; 
    $user=""; 
    $curr_timestamp = now(); 
    define('HOST','********'); 
    define('USER','*******'); 
    define('PASS','*****'); 
    define('DB','********'); 
    $con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect'); 
    $sql = "INSERT INTO 0_users (`ID`, `user_login`, `user_pass`, `user_nicename`, `user_email`, `user_url`, `user_registered`, `user_activation_key`, `user_status`, `display_name`) VALUES (NULL, '$userusname', '$useruspassword', '$usernickname', '$usermail', '', '$curr_timestamp', '', '0', '$userfullname')"; 
    $check = mysqli_fetch_array(mysqli_query($con,$sql)); 
    if(isset($check)){ 
    echo "success"; 
     }else{ 
    echo "Invalid Username or Password"; 
    } 
    mysqli_close($con); 
    }else{ 
    echo "error try again"; 
    } 
    ?>

來源

2016-02-20 Gnana Prakash

可以顯示當你運行這段代碼是什麼值插入。 –

從android發佈的值已被使用我已經使用回聲來檢查值是否傳遞和值發佈到這個PHP頁面。 –

向我們展示這些變量通過爲每個 –

試試這個

<?php 
     if($_SERVER['REQUEST_METHOD']=='POST'){ 
     $userfullname=preg_replace("/\r|\n/", "", $_POST['userfullname']); 
     $usernickname=preg_replace("/\r|\n/", "", $_POST['usernickname']); 
     $usermail=preg_replace("/\r|\n/", "", $_POST['usermail']); 
     $userusname=preg_replace("/\r|\n/", "", $_POST['userusname']); 
     $useruspassword=preg_replace("/\r|\n/", "", $_POST['useruspassword']); 
     $user=""; 
     $curr_timestamp = now(); 
     define('HOST','********'); 
     define('USER','*******'); 
     define('PASS','*****'); 
     define('DB','********'); 
     $con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect'); 
     $sql = "INSERT INTO 0_users (`user_login`, `user_pass`, `user_nicename`, `user_email`, `user_url`, `user_registered`, `user_activation_key`, `user_status`, `display_name`) VALUES ('$userusname', '$useruspassword', '$usernickname', '$usermail', '', '$curr_timestamp', '', '0', '$userfullname')"; 
     $check = mysqli_query($con,$sql); 
     if(isset($check)){ 
     echo "success"; 
      }else{ 
     echo "Invalid Username or Password"; 
     } 
     mysqli_close($con); 
     }else{ 
     echo "error try again"; 
     }

？>

來源

2016-02-20 12:07:12

我已經使用了代碼，因爲你已經提到，但仍然得到相同的錯誤 –

是
即將在您的代碼？ –

否（而不是顯示這些回聲語句）如果（isset（$ check））{echo「success」; } else { echo「無效的用戶名或密碼」; }我在
echo聲明 –

在向mysql插入值時獲取<br />

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