[英]How to use a $result variable with table object in SQL query using mySQLi
我試圖讓這段代碼工作,但它只工作到第二個 echo 語句echo "Finished 2";
.
<?php
if (count($_GET) > 0){
$sql = "SELECT * FROM winery WHERE winery_name='".$_GET['winery_name']."'";
echo "Finished 1";
$result = $db->query($sql);
echo "Finished 2";
$sql = "SELECT * FROM".$result."WHERE wine_type='".$_GET['wine_type']."'";
echo "Finished 3";
$result = $db->query($sql);
echo "Finished 4";
$sql = "SELECT * FROM".$result.", wine_variety WHERE wine_id=wine_variety.wine_id";
echo "Finished 5";
$result = $db->query($sql);
echo "Finished 6";
$sql = "SELECT * FROM".$result."WHERE variety_id='".$_GET['grape_variety']."'";
echo "Finished 7";
$result = $db->query($sql);
echo "Finished all queries";
}
?>
根據我的理解,問題是 sql 無法將$result
識別為表,但$result
存儲了我查詢的返回表。 如何讓 SQL 在新查詢中使用$result
的返回表?
我想從你的酒廠表你正在獲取其他表名???
如果是這樣,您需要從 $result 中獲取行,然后從酒廠表中獲取適當的列(即具有其他表名的列)。
順便說一句,最好的選擇是加入兩張桌子。
我認為你犯錯的另一點是
$sql = "SELECT * FROM".$result."WHERE wine_type='".$_GET['wine_type']."'";
應該
$sql = "SELECT * FROM ".$result." WHERE wine_type='".$_GET['wine_type']."'";
FROM 和雙引號之間以及雙引號和 WHERE 之間的空格
要從 winary_name 獲取 winery_id,您可以編寫 HTML 表單,例如
<select name="winary_id">
<option value="Winary ID HERE">Winary Name Here</option> // you can generate your dynamic options like this which will return id instead of name
</select>
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