Question

我想在PHP中编写一个脚本，它接受页面的当前URL，在数据库中搜索它（带有WAMP的phpMyAdmin）并打印带有该URL的行的ID。我已经写了这个，并且到了连接点，everythwing工作，与mysql的连接也工作正常，问题从这一行开始：

 $query = "SELECT id FROM search WHERE link = '". $curPageURL . "'";

任何人都可以告诉我：我做错了什么？这是整个剧本：

<!DOCTYPE html>
<html>
 <body>

 <?php
 function curPageURL() {
 $pageURL = 'http';
 if ($_SERVER["HTTPS"] == "on") {$pageURL .= "s";}
 $pageURL .= "://";
 if ($_SERVER["SERVER_PORT"] != "80") {
  $pageURL .= $_SERVER["SERVER_NAME"].":".$_SERVER["SERVER_PORT"].$_SERVER["REQUEST_URI"];
 } else {
  $pageURL .= $_SERVER["SERVER_NAME"].$_SERVER["REQUEST_URI"];
 }
 return $pageURL;
 }
 echo curPageURL();

 //connection
 $con= mysql_connect("localhost","root","") or die ("Could not connect");
 mysql_select_db("search") or die ("Could not select db");
 echo "connection succesful";

 $query = "SELECT id FROM search WHERE link = '". $curPageURL . "'";

 while($row = mysql_fetch_array($query)) {
 echo $row;
 }
 ?>

 </body>
 </html>

Answer 1

您没有使用mysql_query()运行查询，也没有设置变量$curPageURL。

<!DOCTYPE html>
<html>
 <body>

 <?php
 function curPageURL() {
 $pageURL = 'http';
 if ($_SERVER["HTTPS"] == "on") {$pageURL .= "s";}
 $pageURL .= "://";
 if ($_SERVER["SERVER_PORT"] != "80") {
  $pageURL .= $_SERVER["SERVER_NAME"].":".$_SERVER["SERVER_PORT"].$_SERVER["REQUEST_URI"];
 } else {
  $pageURL .= $_SERVER["SERVER_NAME"].$_SERVER["REQUEST_URI"];
 }
 return $pageURL;
 }
 echo curPageURL();

 //connection
 $con= mysql_connect("localhost","root","") or die ("Could not connect");
 mysql_select_db("search") or die ("Could not select db");
 echo "connection succesful";

 $query = "SELECT id FROM search WHERE link = '". curPageURL() . "'";
 $result = mysql_query($query);

 while($row = mysql_fetch_array($result)) {
 echo $row;
 }
 ?>

 </body>
 </html>

请注意，您使用的方法已从PHP 5.5.0中弃用。所以我建议你考虑mysqli或PDO。示例可以在下面的php手册链接中找到

http://www.php.net/manual/en/mysqli.query.php

http://www.php.net/manual/en/pdo.query.php

Answer 2

您可能想尝试这样的事情。

$query = "SELECT id FROM search WHERE link = '". curPageURL() . "'";
$result = mysql_query($query);

while($row = mysql_fetch_array($result)) {
    echo $row['id'];
}

我无法在我的PHP脚本中找到错误

2 个答案: