错误处理php_network_getaddresses：getaddrinfo失败。怎么样？

Question

我想抛出一个异常，并捕获一个错误，然后回显一个数字。

这是我到目前为止，但它返回以下PHP错误，从而使我的AJAX beforeSend请求挂起。

PHP代码

<?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 */ 

    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"]; 

    if(!$link = mysqli_connect("intentionally_mispelled", "correct_user", "correct_pass", "correct_database")) { 
    echo "2"; 
    exit; 

    } else { 
    /* 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); 

    } 
    } 

} 

?> 

如何能为我上述我优雅地处理这个问题？

<br /> 
    <b>Warning</b>: mysqli_connect(): php_network_getaddresses: getaddrinfo failed: Name or service not known in <b>/nfs/c12/h02/mnt/220474/domains/site.com/html/autosave4/processor.php</b> on line <b>20</b><br /> 
    <br /> 
    <b>Fatal error</b>: Uncaught exception 'mysqli_sql_exception' with message 'php_network_getaddresses: getaddrinfo failed: Name or service not known' in /nfs/c12/h02/mnt/220474/domains/nexlevel.org/html/autosave4/processor.php:20 
    Stack trace: 
    #0 /nfs/c12/h02/mnt/220474/domains/nexlevel.org/html/autosave4/processor.php(20): mysqli_connect('intentionally_mispelled', 'correct_username', 'correct_password', 'correct_database') 
    #1 {main} 
     thrown in <b>/nfs/c12/h02/mnt/220474/domains/site.com/html/autosave4/processor.php</b> on line <b>20</b><br /> 

基本上我只是想回应一个简单的数字2，但PHP打断了我的回声和输出低于这个错误堆栈。

最终代码 - 工作& COMMENTED

<?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 */ 

    /* Sample URL */ 
    // mysite.org/autosave5/processor.php?postCLIENT_ID=111&postPROJECT_ID=222&postMAINSHEET_ID=333&postT_ID=WTRESRVD&postT_VAL=147 

    /* Disable Warnings so that we can return ONLY what we want through echo. */ 
    mysqli_report(MYSQLI_REPORT_STRICT); 

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

    // Initialize Global variables. 
    $client_id = ''; 
    $project_id = ''; 
    $mainsheet_id = ''; 
    $field_name = ''; 
    $field_value = ''; 

    /* Database Connection Check */ 
    try 
    { 
    if ($link = mysqli_connect("incorrect_domain", "correct_username", "correct_password", "correct_database")) 
    { 

     // Set and Escape Global variables. 
     $client_id = mysqli_real_escape_string($link, $_GET["postCLIENT_ID"]); 
     $project_id = mysqli_real_escape_string($link, $_GET["postPROJECT_ID"]); 
     $mainsheet_id = mysqli_real_escape_string($link, $_GET["postMAINSHEET_ID"]); 
     $field_name = mysqli_real_escape_string($link, $_GET["postT_ID"]); 
     $field_value = mysqli_real_escape_string($link, $_GET["postT_VAL"]); 

     /* 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, $sql)) { 
     /* return 3 = MySQL Query Error OR Wrong URL Parameters */ 
     echo "3"; 
     /* Close Connection */ 
     mysqli_close($link); 
     exit; 
     } else { 
     /* return 0 = Nothing to Update/1 = Successful Update Query */ 
     echo "".mysqli_affected_rows($link); 
     /* Close Connection */ 
     mysqli_close($link); 
    } 

    } else { 
     throw new Exception('2'); 
    } 
    } catch(Exception $e) { 
    /* echo $e->getMessage(); 
     return 2 = Database Connection refused */ 
    echo "2"; 
    } 

} 

?> 

Answer 1

如果你想赶上并抛出exceptionm不要使用这种if声明。使用try ... catch

try 
{ 
    if ($db = mysqli_connect($hostname_db, $username_db, $password_db, $base)) 
    { 
    //do something 
    } 
    else 
    { 
     throw new Exception('Unable to connect'); 
    } 
} 
catch(Exception $e) 
{ 
    echo $e->getMessage(); 
}