我新的PHP和我在我的網址作爲參考使用ID試圖從我的DB呼應某一特定領域從MySQL呼應。但是,不要回顯它回顯ID的值。PHP-有問題,使用GET
我究竟做錯了什麼?有沒有更好的方法來做到這一點?提前致謝。
我一個試圖從我在我的網址ID呼應dateCreated會現場。 URL是www.example.com/client_profile.php?id=1001
<div class="thumb-info mb-md">
<img src="assets/images/!logged-user.jpg" class="rounded img-responsive" alt="John Doe">
<div class="thumb-info-title">
<?php
echo $_GET['id']; //Output: myquery
$result = mysql_query("SELECT * FROM client WHERE id = '" . $_GET['id'] . "'");
while($row = mysql_fetch_array($result)){
echo "<span class='thumb-info-inner'>$row.['datecreated'].</span>";
}
?>
</div>
</div>
更新的代碼
<?php
mysql_connect("localhost", "my user", "my pass") or
die("Could not connect: " . mysql_error());
mysql_select_db("my db");
$id = (int)$_GET['id']; //Output: myquery
echo $id;
$result = mysql_query("SELECT * FROM client WHERE id = $id");
while($row = mysql_fetch_array($result)){
echo "<span class='thumb-info-inner'>" . $row['datecreated'] . "</span>";
}?>
呼應的兩個編號,並從DB
,您在該線路上'回聲說 「<跨度類= '拇指信息-內'> $行[ 'dateCreated會']。」;}?>'它輸出1001? – Leeish
@Leeish是標識beeing 1001 – Mathew
'1001'從'回聲的$ id來;'。沒有辦法來自'$ row ['datecreated']' – Barmar