mysql_fetch_assoc（）期望参数1是资源，在第9行中给出的布尔值为C：\xampp\htdocs\qcc\truckdelivery.php没有选择数据库 [英] mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\qcc\truckdelivery.php on line 9 no database selected
问题描述
<?php require_once('Connections/connect.php'); ?>
<?php
$id_truck = mysql_real_escape_string($_GET['id_truck']);
$sql_PK = "SELECT * FROM tbl_truck WHERE id_truck = '{$id_truck}'";
$res_PK = mysql_query($sql_PK, $connect);
$row_PK = mysql_fetch_assoc($res_PK);
$truck_plate = $row_PK['truck_plate_no'];
$truck_plate = mysql_real_escape_string($truck_plate);
$sql = "SELECT tbl_truck.truck_plate_no,
tbl_delivery_details.delivery_details_route,
tbl_delivery_details.delivery_details_destination,
tbl_delivery_details.delivery_details_van_no,
tbl_delivery_details.delivery_details_waybill_no,
tbl_delivery_details.delivery_details_charge_invoice,
tbl_delivery_details.delivery_details_revenue,
tbl_delivery_details.delivery_details_strip_stuff,
tbl_delivery_details.delivery_details_date
FROM tbl_truck, tbl_delivery_details
WHERE tbl_truck.truck_plate_no = tbl_delivery_details.tbl_truck_id_truck
AND tbl_truck.truck_plate_no = '{$truck_plate}'";
$res = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($res);
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Truck Delivery</title>
<link rel="stylesheet" type="text/css" href="qcc.css"/>
</head>
<body>
<?php echo $row_PK['delivery_details_route'];?>
</body>
</html>
mysql_fetch_assoc（）期望参数1为资源，布尔值为C ：\xampp\htdocs\qcc\truckdelivery.php on line 9
没有选择数据库
mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\qcc\truckdelivery.php on line 9 no database selected
I don不知道我的代码错在哪里;这个页面是一个表格中的一个链接，我使用这个代码将板号编成链接。
I don't know where I got the code wrong; this page is a link from a table that I made full of plate numbers, and I made the plate number into a link using this code.
href="truckdelivery.php?id_truck=<?php echo $row_truck['id_truck']?>"><?php echo $row_truck['truck_plate_no'];
推荐答案
在处理查询结果之前检查MySQL错误。很可能您的查询出错了，因此 mysql_query（）没有返回有效结果，因此 mysql_fetch_assoc（）失败。
Check for MySQL errors before processing the result of a query. Most likely your query went wrong some way, so mysql_query() did not return a valid result and thus mysql_fetch_assoc() fails.
// ...
$PK = mysql_query($sql_PK, $connect);
if ( mysql_error() ) {
die ( mysql_error();
}
$row_PK = mysql_fetch_assoc($PK);
// ...
除此之外： mysql_x（）函数已弃用。使用 PDO 或 MySQLi 来代替。
Besides: mysql_x() functions are deprecated. Use PDO or MySQLi instead.
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