我想通过url传递多个值，如下所示。
http://example.com/shopping_cart_json_api.php?uid=5710，55
我已经创建了php编码文件如下，但我无法得到正确的结果。
<?php
require('adminpanel/includes/application_top.php');
if ($_SERVER['REQUEST_METHOD'] == 'GET') {
$id = explode(",", $_GET["uid"]);
$sth = mysql_query(" SELECT drug_medicines.fld_id,
drug_medicines.fld_image, drug_medicines.ld_product_name,
drug_medicines.fld_best_price,drug_cart.ld_userid,
drug_cart.fld_qunty, drug_cart.fld_totalprice
FROM drug_medicines
INNER JOIN drug_cart
ON drug_medicines.fld_id=drug_cart.fld_productid
WHERE fld_userid='" . $id . "' ||'" . $id . "'");
$rows = array();
while ($r = mysql_fetch_assoc($sth)) {
$rows[] = $r;
}
print json_encode($rows);
}
?>
格式应该是这样的。
WHERE fld_userid='".$id[0]."' OR fld_userid= '".$id[1]."'");
对于大于2的值，使用
WHERE fld_userid IN (" . implode(',', $id) . ")"
根据您的URL，您将来可能会有更多的值，因此您应该使用mysql 的IN函数
fld_userid IN ($id);
在mysql查询中使用IN。
<?php
require('adminpanel/includes/application_top.php');
if ($_SERVER['REQUEST_METHOD'] == 'GET') {
$id = explode(",", $_GET["uid"]);
$sth = mysql_query("
SELECT
drug_medicines.fld_id,
drug_medicines.fld_image, drug_medicines.ld_product_name,
drug_medicines.fld_best_price,drug_cart.ld_userid,
drug_cart.fld_qunty, drug_cart.fld_totalprice
FROM
drug_medicines
INNER JOIN
drug_cart
ON
drug_medicines.fld_id=drug_cart.fld_productid
WHERE
fld_userid IN (" . $_GET["uid"] . ")");
$rows = array();
while ($r = mysql_fetch_assoc($sth)) {
$rows[] = $r;
}
print json_encode($rows);
}
?>
您应该使用IN以获得更大的灵活性。
"... WHERE fld_userid IN (" . implode(',', $id) . ")"
将输出：
"... WHERE fld_userid IN (5710,55)"