輸出新的查詢來自晶

Question

我試圖找出如何讓改變查詢 示例（HTML）

<form action="change.php" method="POST" name="Update"> 
<table> 
    <tr> 
    <td> 
     <input type="text" value="Enter New Criteria" name="where" > 
     <input type="text" value="Enter New Criteria" name="where2" > 
    </td> 
    </tr> 
    <tr> 
    <td align="center" style="font-family:Calibri"> 
    <input type="submit" value="Search"/> 
    </tr> 
</table> 

SQL查詢形式

$Query = "SELECT order_number 
      FROM order_header 
      WHERE (order_number LIKE **'%CHANGE VALUE HERE%'** 
      OR order_number LIKE **'%CHANGE VALUE HERE%'** 

我怎麼會去這樣做那我是一個完整的菜鳥，但我在努力。我試過搜索，但也許我沒有使用正確的關鍵詞。

Answer 1

使用$_POST['where']和$_POST['where2']代替%CHANGE VALUE HERE%。更多信息：http://php.net/manual/en/reserved.variables.post.php

例如：

$Query = "SELECT order_number 
      FROM order_header 
      WHERE (order_number LIKE '$_POST[where]' 
      OR order_number LIKE '$_POST[where2]')"; 

當你得到它的工作，有一個關於SQL注入讀：http://php.net/manual/en/security.database.sql-injection.php

Answer 2

嘗試：

$where = $_POST['where']; 
$where2 = $_POST['where2']; 

$Query = "SELECT order_number 
      FROM order_header 
      WHERE (order_number LIKE '%whereParameter%' 
      OR order_number LIKE '%where2Partameter%'" 

$Query = str_replace("whereParameter", $where, $QueryTemplate); 
$Query = str_replace("where2Parameter", $where2, $QueryTemplate);