我已經寫了一段代碼,但我似乎無法使它工作。我確實與我的數據庫有聯繫。當我提交表格時,它會轉到else
區塊(echo "Geen resultaat gevonden voor \"<b>$s</b>\"";
)爲什麼我的PHP搜索引擎不能找到任何記錄?
我在做什麼錯我的代碼?我還添加了我的數據庫的屏幕截圖。
<body>
<h2> hier komt een kleine foto</h2>
<form action='./search.php' method='get'>
<input type='text' name='s'size='50' value='<?php echo $_GET['s']; ?>' />
<input type='submit' value='Zoek'/>
</form>
<hr />
<?php
$s = $_GET['s'];
$terms = explode (" ", $s);
$query = "SELECT * FROM 'ID' WHERE ";
foreach ($terms as $each){
$i++;
if ($i ==1)
$query .= "keywords LIKE '%$each%' ";
else
$query .= "OR keywords LIKE '%$each%' ";
}
//connect to database
mysql_connect("server", "username", "password");
mysql_select_db("database");
$query = mysql_query($query);
$numrows = mysql_num_rows($query);
if(numrows > 0){
while($row = mysql_fetch_assoc($query)){
$id = $row['id'];
$photo = $row['photo'];
$title = $row['title'];
$description = $row['description'];
$price = $row['price'];
$Link = $row['Link'];
$keywords = $row['keywords'];
echo "<h2><a href='$Link'>$title</a></h2>
$description<br /><br />";
}
}
else
echo "Geen resultaat gevonden voor \"<b>$s</b>\"";
//disconect
mysql_close();
?>
</body>
</html>
這對我的眼睛來說很痛苦。你很容易[sql注入,xss,] –
爲什麼?你可以幫我請我與這個新的 –
沒關係,現在不用擔心 –