如何在PHP中爲不同的表發送兩個查詢？

我想在名爲「preise」的talbe中創建新數據。如果我這樣做，我想在名爲「tische」的表格中添加一列，並在「preise」中創建數據的標題我希望你能理解我的意思（我的英文不是很好「）這是我的PHP：如何在PHP中爲不同的表發送兩個查詢？

<?php 
    if($_SERVER['REQUEST_METHOD']=='POST'){ 

     //Getting values 
     $name = $_POST['name']; 
     $ebene = $_POST['ebene']; 
     $preis = $_POST['preis']; 

     //Creating an sql query 
     $sql = "INSERT INTO preise (name,ebene,preis) VALUES ('$name','$ebene','$preis')"; 
    $add = "ALTER TABLE tische ADD COLUMN $name INT(5) NOT NULL AFTER id"; 

     //Importing our db connection script 
     require_once('dbConnect.php'); 

     //Executing query to database 
     if(mysqli_query($con,$sql,$add)){ 
      echo 'Element erfolgreich hinzugefügt'; 
     }else{ 
      echo 'Konnte Element nicht hinzufügen'; 
     } 

     //Closing the database 
     mysqli_close($con); 
    }

如果我刪除在$行添加...那麼腳本將執行$ SQL，因爲它應該，但我怎麼能執行$添加查詢

來源

2016-11-06 user3713946

？

不同查詢應分開執行

一個簡單的例子：

//Creating an sql query 
$insert_query = "INSERT INTO preise (name,ebene,preis) VALUES ('$name','$ebene','$preis')"; 
$alter_query = "ALTER TABLE tische ADD COLUMN $name INT(5) NOT NULL AFTER id"; 

//Importing our db connection script 
require_once('dbConnect.php'); 

//Executing query to database 

// mysqli_query takes TWO or THREE arguments only 
// and third argument is an INT value 
// so passing another query as third argument is USELESS 
if (mysqli_query($con, $insert_query)){ 

    // if first query is okay 
    // we can run second one 
    $r = mysqli_query($con, $alter_query); 
    if ($r) { 
     // both queries are OK 
     echo 'Element erfolgreich hinzugefügt'; 
    } else { 
     echo 'Some error message here'; 
    } 
} else { 
    echo 'Konnte Element nicht hinzufügen'; 
} 

//Closing the database 
mysqli_close($con);

如需進一步讀數：

來源

2016-11-06 17:00:08

如何在PHP中爲不同的表發送兩個查詢？

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