首先我寫數據到一個表的「ID」主鍵(作品) 寫入到數據庫在幾秒鐘之文件得到處理,這看起來像:mysql_fetch_array()期望參數1是資源,boolean!
<?php
require_once 'connection.php';
$pName = $_POST['scriptName'];
$pCode = $_POST['code'];
$group = $_POST['group'];
$descri = $_POST['descr'];
$lang = $_POST['lang'];
$date = date("Y-m-d");
$updateq = "INSERT INTO code_tb (id, Script_Name, Description, Code, Language, Date) VALUES ('NULL', '$pName','$descri', '$pCode', '$lang', '$date')";
$result=mysql_query($updateq);
$query = "Select * FROM site_content";
$resultw = mysql_query($query);
header("refresh: 3; ../highlight/test.php?id=$posts[id]");
echo "Succesfully added.";
?>
你看後3秒我們得到轉移到一個網址(看標題)將其傳送和一切,但它不發出後,「ID =」的克林克看起來像
../highlight/test.php一個id? ID = 但它應該是 ../highlight/test.php?id=1(或2或我們)
的test.php的樣子:
<?php
require_once '../includes/connection.php';
$query = "SELECT * FROM site_content WHERE ID = $_GET[id]";
$result = mysql_query($query);
$post = mysql_fetch_array($result);
?>
,我把數據庫內容的網頁上:
<?php echo $post['Script_Name']; ?>
SUM1知道概率?
我是有點新的PHP,ü可以解釋更詳細的,怎麼牛逼解決? –
@Fabian:你正試圖在你的字符串中使用一個名爲'$ posts'的變量,而不是先存儲任何東西。這對你來說足夠詳細嗎? –