PHP MySQLi連接和查詢錯誤處理。怎麼樣？

以下PHP代碼有效。我似乎無法以自定義的方式處理它的錯誤。PHP MySQLi連接和查詢錯誤處理。怎麼樣？

例如，當我故意在連接字符串中拼錯任何返回碼「3」爲數據庫下來，我的AJAX腳本只是掛在beforeSend永遠...

這是我得到：

<?php 

    if(isset($_POST["postT_VAL"])) { 

    $client_id = $_POST["postCLIENT_ID"]; 
    $project_id = $_POST["postPROJECT_ID"]; 
    $mainsheet_id = $_POST["postMAINSHEET_ID"]; 
    $field_name = $_POST["postT_ID"]; 
    $field_value = $_POST["postT_VAL"]; 

    $link = mysqli_connect("database.domain.com", "username1", "password1", "db220474"); 

    if (!$link) { 

    /* return 3 = database offline */ 
    echo "3"; 

    } else { 

    /* build query */ 
    $sql = "UPDATE tbl_mainsheet2 SET ".$field_name." = '".$field_value."' WHERE client_id = '".$client_id."' AND project_id = '".$project_id."' AND mainsheet_id = '".$mainsheet_id."'"; 

    /* execute query */  
    mysqli_query($link, $sql); 

    /* return 0 = no update/1 = successful update */ 
    echo "".mysqli_affected_rows($link); 

    /* close connection */ 
    mysqli_close($link); 

} 

} 

?>

新的研究

好吧。經過一番研究後，我發現這是有效的。看來你需要告訴mysqli拋出異常。由於某種原因，這不同於僅僅嘗試處理「IF」錯誤的方式。任何改進建議？

mysqli_report(MYSQLI_REPORT_STRICT); 

     try { 
      $link = mysqli_connect("database.domain.com", "username1", "password1", "db220474"); 
     } catch (Exception $e) { 
      echo "3"; 
      exit; 
     }

代碼更新

這裏是有目共睹的最後測試和工作PHP的解決方案。

<?php 

    /* Status Codes 

    return 0 = Nothing to Update 
    return 1 = Successful Update Query 
    return 2 = Database Connection refused 
    return 3 = MySQL Query Error OR Wrong URL Parameters */ 

    mysqli_report(MYSQLI_REPORT_OFF); 

    if(isset($_GET["postT_VAL"])) { 

    $client_id = $_GET["postCLIENT_ID"]; 
    $project_id = $_GET["postPROJECT_ID"]; 
    $mainsheet_id = $_GET["postMAINSHEET_ID"]; 
    $field_name = $_GET["postT_ID"]; 
    $field_value = $_GET["postT_VAL"]; 

    try { 
     $link = mysqli_connect("domain", "username", "password", "database"); 
    } catch (Exception $e) { 
     // echo "".$e->getCode(); 
     /* return 2 = Database Connection refused */ 
     echo "2"; 

     exit; 
    } 

    /* Build dynamic Update Query string */ 
    $sql = "UPDATE tbl_mainsheet2 SET ".$field_name." = '".$field_value."' WHERE client_id = '".$client_id."' AND project_id = '".$project_id."' AND mainsheet_id = '".$mainsheet_id."'"; 

    /* Execute Update Query */  
    if(!mysqli_query($link, $sql)) { 

    echo "3"; 
    /* Close Connection */ 
    mysqli_close($link); 

    exit; 

    } else { 

    /* return 0 = Nothing to Update/1 = Successful Update Query */ 
    echo "".mysqli_affected_rows($link); 

    /* Close Connection */ 
    mysqli_close($link); 

    } 

} 

?>

來源

2017-08-03 ASPiRE

使用'mysqli_error（$鏈接）'揭示錯誤。此外，您的代碼易受SQL注入攻擊。請檢查並重寫。 – Raptor

如果你只是在你的if內運行代碼，那麼你可能會得到一個「Ahha」時刻，即將該部分複製到一個新文件中 - 或者註釋掉現有if和直接運行頁面... – TimBrownlaw

'echo'3 「'你需要添加'exit（）'或'die（$ message）'來獲取錯誤信息，並且如果你的連接有錯誤，就不要執行下一行代碼。 –

試試這個在執行查詢

/* execute query */  
    mysqli_query($link, $sql) or die(mysqli_error($link));

來源

2017-08-03 03:36:26

enter code here 
<?php 
$con=mysqli_connect("localhost","my_user","my_password","my_db"); 

// Check connection 
if (mysqli_connect_errno()) 
{ 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

// Perform a query, check for error 
if (!mysqli_query($con,"INSERT INTO Persons (FirstName) VALUES ('Glenn')")) 
{ 
echo("Error description: " . mysqli_error($con)); 
} 
mysqli_close($con); 
?>

試試這個好運氣

來源

2017-08-03 03:51:35 Jok3r

這裏是有目共睹的最後測試和工作PHP的解決方案。

<?php 

    /* Status Codes 

    return 0 = Nothing to Update 
    return 1 = Successful Update Query 
    return 2 = Database Connection refused 
    return 3 = MySQL Query Error OR Wrong URL Parameters */ 

    mysqli_report(MYSQLI_REPORT_STRICT); 

    if(isset($_GET["postT_VAL"])) { 

    $client_id = $_GET["postCLIENT_ID"]; 
    $project_id = $_GET["postPROJECT_ID"]; 
    $mainsheet_id = $_GET["postMAINSHEET_ID"]; 
    $field_name = $_GET["postT_ID"]; 
    $field_value = $_GET["postT_VAL"]; 

    try { 
     $link = mysqli_connect("domain", "username", "password", "database"); 
    } catch (Exception $e) { 
     // echo "".$e->getCode(); 
     /* return 2 = Database Connection refused */ 
     echo "2"; 

     exit; 
    } 

    /* Build dynamic Update Query string */ 
    $sql = "UPDATE tbl_mainsheet2 SET ".$field_name." = '".$field_value."' WHERE client_id = '".$client_id."' AND project_id = '".$project_id."' AND mainsheet_id = '".$mainsheet_id."'"; 

    /* Execute Update Query */  
    if(!mysqli_query($link, $sql)) { 

    echo "3"; 
    /* Close Connection */ 
    mysqli_close($link); 

    exit; 

    } else { 

    /* return 0 = Nothing to Update/1 = Successful Update Query */ 
    echo "".mysqli_affected_rows($link); 

    /* Close Connection */ 
    mysqli_close($link); 

    } 

} 

?>

來源

2017-08-03 12:06:54 ASPiRE

我還沒有能夠在這裏找到解決方案，除了查詢的錯誤處理之外，我還沒有找到任何答案。 – ASPiRE

你可以這樣做：

try { 
    $link = mysqli_connect("domain", "username", "password","database"); 
} catch (Exception $e) { 
     echo $e; 
    exit; 
}

來源

2017-08-03 12:26:06

PHP MySQLi連接和查詢錯誤處理。怎麼樣？

回答

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