PHP MySQLi连接和查询错误处理。怎么样？

以下PHP代码有效。我似乎无法以自定义的方式处理它的错误。PHP MySQLi连接和查询错误处理。怎么样？

例如，当我故意在连接字符串中拼错任何返回码“3”为数据库下来，我的AJAX脚本只是挂在beforeSend永远...

这是我得到：

<?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。经过一番研究后，我发现这是有效的。看来你需要告诉mysqli抛出异常。由于某种原因，这不同于仅仅尝试处理“IF”错误的方式。任何改进建议？

mysqli_report(MYSQLI_REPORT_STRICT); 

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

代码更新

这里是有目共睹的最后测试和工作PHP的解决方案。

<?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); 

    } 

} 

?>

来源

2017-08-03 ASPiRE

使用'mysqli_error（$链接）'揭示错误。此外，您的代码易受SQL注入攻击。请检查并重写。 – Raptor

如果你只是在你的if内运行代码，那么你可能会得到一个“Ahha”时刻，即将该部分复制到一个新文件中 - 或者注释掉现有if和直接运行页面... – TimBrownlaw

'echo'3 “'你需要添加'exit（）'或'die（$ message）'来获取错误信息，并且如果你的连接有错误，就不要执行下一行代码。 –

试试这个在执行查询

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

来源

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); 
?>

试试这个好运气

来源

2017-08-03 03:51:35 Jok3r

这里是有目共睹的最后测试和工作PHP的解决方案。

<?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); 

    } 

} 

?>

来源

2017-08-03 12:06:54 ASPiRE

我还没有能够在这里找到解决方案，除了查询的错误处理之外，我还没有找到任何答案。 – ASPiRE

你可以这样做：

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

来源

2017-08-03 12:26:06

PHP MySQLi连接和查询错误处理。怎么样？

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