Question

只是尝试使用php＆＃34; echo＆＃34;来粘贴一些值。并且我运气不好。

我相信我使用的代码在其第一个变量设置＆＃34; $ theid＆＃34;中有问题，它从表中获取id字段。这是代码：

<?php

$conn=mysql_connect("localhost", "username", "password")
or die ("Cannot Connect to MYSQL");

mysql_select_db("database1", $conn)
or die ("Cannot Connect to the Database");
echo "Connected successfully";

$theid = isset($_GET['id'])?$_GET['id']:""; //Possible problematic code

$data = mysql_fetch_array(mysql_query("SELECT * FROM table1 WHERE id='$theid'"));

?>

然后将使用数据：

<?php echo $data['url'] ?>

问题是，＆＃34; $ data＆＃34;

下没有显示任何内容

在对此进行故障排除并环顾SO之后，我仍然没有找到答案。任何反馈都非常感谢，我确信这只是我使用语法时出错！感谢。

Answer 1

使用while循环打印查询结果数据

<?php

$conn=mysql_connect("localhost", "username", "password")
or die ("Cannot Connect to MYSQL");

mysql_select_db("database1", $conn)
or die ("Cannot Connect to the Database");
echo "Connected successfully";

if(isset($_GET['id'])){
   $theid = $_GET['id'];
}

$result = mysql_query("SELECT * FROM table1 WHERE id='$theid'");

while($data = mysql_fetch_array($result)){
    echo $data['url'];
}

?>

Answer 2

$conn=mysql_connect("localhost", "username", "password")
or die ("Cannot Connect to MYSQL");

mysql_select_db("database1", $conn)
or die ("Cannot Connect to the Database");
echo "Connected successfully";

//default value
$theid = 0;
if(isset($_GET['id'])){
   $theid = $_GET['id'];
}

$result = mysql_query("SELECT * FROM table1 WHERE id='".(int)$theid."'");

while($data = mysql_fetch_array($result)){
    echo $data['url'];
}

转换为int以防止SQL注入，并且var不能被解释为简单引用或不使用＆＃39;＆＃39; 在PHP中，var被解释为双引号。

Answer 3

$result = mysql_query("SELECT * FROM table2 WHERE id='$theid'");

while($data = mysql_fetch_array($result)){
    echo $data['url'];

错误地命名表

$ _GET [＆＃39; id＆＃39;]检索数据时出现问题

3 个答案: