PDO提取設置變量mysql php

Question

我是新來的這個PDO與PHP的東西，所以我想知道如何將數據分配給變量。

當它返回時我得到這個Array ([0] => Array ([id] => 1 [title] => Test Announcement! [link] => http://www.google.com))。

我希望它只是「測試公告」和「http://www.google.com」，並且能夠將它們簽名爲像$ title和$ link這樣的變量。

的functions.php

<?php 
    require("common.php"); 
    function getAnnouncements() { 
     $query = "SELECT title, link FROM announcements"; 

     try { 
      global $db; 
      // Execute the query against the database 
      $stmt = $db->prepare($query); 
      $stmt->execute(); 
      $result = $stmt->fetchAll(); 
      print_r($result); 

     } 
     catch(PDOException $ex) { 
      // Note: On a production website, you should not output $ex->getMessage(). 
      // It may provide an attacker with helpful information about your code. 
      die("Error loading announcements"); 
     } 

    } 

?> 

我用kypros回答，現在我有這個：

<?php 
    require("common.php"); 
    function getAnnouncements() { 
     $query = "SELECT title, link FROM announcements"; 

     try { 
      global $db; 
      // Execute the query against the database 
      $stmt = $db->prepare($query); 
      $stmt->execute(); 
      $result = $stmt->fetchAll(); 
      foreach($result as $announcement) 
       $title = $announcement['title']; 
       $link = $announcement['link']; 
       echo "<a href='".$link."'>".$title."</a>"; 
      } 
     } 
     catch(PDOException $ex) { 
      // Note: On a production website, you should not output $ex->getMessage(). 
      // It may provide an attacker with helpful information about your code. 
      die("Error loading announcements"); 
     } 

    } 

?> 

隨着error.log中和空白頁。

[18-Oct-2014 13:28:12 UTC] PHP Parse error: syntax error, unexpected '}', expecting T_CATCH in /home/dxbridge/public_html/scripts/functions.php on line 17 
[18-Oct-2014 13:28:13 UTC] PHP Parse error: syntax error, unexpected '}', expecting T_CATCH in /home/dxbridge/public_html/scripts/functions.php on line 17 

Answer 1

您的$result現在包含與您的select語句匹配的所有行。要訪問各個列值，你可以這樣做：

foreach($result as $announcement) 
{ 
    $title = $announcement['title']; 
    $link = $announcement['link']; 
    echo "<a href='".$link."'>".$title."</a>"; 
} 

這將輸出連接每個公告的鏈接並顯示超鏈接只是它的標題