Question

如何檢查使用的查詢是更新我的桌子做或不使用PHP

下面

是我的代碼

$sql_query = "update companies set 
     first_name = '$first_name', 
     last_name = '$last_name', 
     designation = '$designation', 
     company_name = '$company_name', 
     street_address = '$street_address', 
     city_code = '$city_code', 
     telephone_number = '$telephone_number', 
     mobile_number = '$mobile_number', 
     fax_number = '$fax_number' 
     where company_code='1001';"; 

    if (!mysqli_query($conn_1,$sql_query)) 
    { 
     $_SESSION['error_details'][0] = 'no'; 
     $_SESSION['error_details'][1] = 'Sorry, not update!'; 
    } 
    else 
    { 
     $_SESSION['error_details'][0] = 'yes'; 
     $_SESSION['error_details'][1] = 'Thank you, update sucessfully!'; 
    } 

如何「如果條件」檢查此更新

Answer 1

使用mysqli.affected-rows檢查更新的查詢進行任何更新或沒有。

if(mysqli_affected_rows($con) > 0) { 
    //update performed 
} 

Answer 2

返回受上一INSERT，UPDATE，行數更換或DELETE查詢。

mysqli_affected_rows()將返回受更新影響的行數。

http://www.php.net/manual/en/mysqli.affected-rows.php

Answer 3

PHP如果你正在更新的數據是從你的數據庫中的數據不同，一個好的做法是使用一類或ORM，如果你喜歡純PHP，用類或函數照顧只是將更新正在做的事情就像使用ORM或創建一個類來抽象的數據庫連接

Answer 4

的mysqli_affected_rows（）本功能（照顧錯誤將被顯示在立法院的開發時間）

$result = mysql_query($sqlString) or die(mysql_eror()); // then you will be sure there is no errors; 

if (mysql_affected_rows ($result) <= 0) { 
//do as you need for no changes on your database 
} else { 
//do as you want on database changed 
} 

不過是更好ion返回以前的SELECT，INSERT，UPDATE，REPLACE或DELETE查詢中受影響的行數。

瞭解更多：

http://php.net/manual/en/mysqli.affected-rows.php

$conn_1 = mysqli_connect("localhost", "DB_USER", "DB_PASSWORD", "DB_NAME"); 

    $sql_query = "update companies set 
     first_name = '$first_name', 
     last_name = '$last_name', 
     designation = '$designation', 
     company_name = '$company_name', 
     street_address = '$street_address', 
     city_code = '$city_code', 
     telephone_number = '$telephone_number', 
     mobile_number = '$mobile_number', 
     fax_number = '$fax_number' 
     where company_code='1001'"; 

    mysqli_query($conn_1,$sql_query); 

    if (mysqli_affected_rows($conn_1)) 
    { 

     $_SESSION['error_details'][0] = 'yes'; 
     $_SESSION['error_details'][1] = 'Thank you, update sucessfully!'; 
    } 
    else 
    { 
     $_SESSION['error_details'][0] = 'no'; 
     $_SESSION['error_details'][1] = 'Sorry, not update!'; 
    }