Question

任何人都可以看到我如何使用String而不是Numeral用于此语句，

我试图在$ delivery =“ALL”时用$ delivery拉出所有行但是我只能拉1个单一变量，如果我想拉数组，我不能，因为我得到数组转换数为字符串，

$delivery_con = [];
if ($delivery==="collection") { $delivery_con[]="no";}
if ($delivery==="delivery") {$delivery_con[]="yes";}
if ($delivery==="local") {$delivery_con[]="yes local";}

if ($delivery==="either") {$delivery_con=["no","yes","yes local"];}


$query="SELECT * 
        FROM testdata 
        WHERE title LIKE ? 
        AND location LIKE ? 
        AND postcode LIKE ? 
        AND price >=? 
        AND price <=? 
        AND cond=? 
        AND catagory LIKE ? 
        AND delivery IN ? 
        ORDER BY $order $dir";

$stat=$db->prepare($query);

$stat->execute(array("%$searchfor%",
                    "%$location%",
                    "%$postcode%",
                    "$pricefrom",
                    "$priceto",
                    "$cond",
                    "%$catagory%",
                    "$delivery_con"));

所以我的问题是，如何解决这个问题，使select函数与$ variables一起使用，

我真的被困住了。如果有人可以提供帮助

谢谢。

Answer 1

使用IN（）SQL运算符时，需要手动创建一组?并将它们放入查询中：

<?php

$delivery_con = [];

if ($delivery === "collection") {
    $delivery_con[] = "no";
}
if ($delivery === "delivery") {
    $delivery_con[] = "yes";
}
if ($delivery === "local") {
    $delivery_con[] = "yes local";
}

if ($delivery==="either") {$delivery_con=["no","yes","yes local"];}


$in = str_repeat('?,', count($delivery_con) - 1) . '?';


$searchfor = "%$searchfor%";
$location  = "%$location%";
$postcode  = "%$postcode%";
$catagory  = "%$catagory%";

$array1 = array($searchfor,$location,$postcode,$pricefrom,$priceto,$cond,$catagory);

$params = array_merge($array1, $delivery_con);

$query = "SELECT * 
        FROM testdata 
        WHERE title LIKE ? 
        AND location LIKE ? 
        AND postcode LIKE ? 
        AND price >=? 
        AND price <=? 
        AND cond=? 
        AND catagory LIKE ? 
        AND delivery IN ($in) 
        ORDER BY $order $dir";

$stat = $db->prepare($query);

$stat->execute([$params]);

Mysql SELECT WHERE $ VAR IN $ VAR，$ VAR，$ VAR

1 个答案: