我已经在sql中编写了代码,因为我只想了解这个概念.完成后我会将其更改为sqli / pdo.所以请忽略sql方法,并帮助我解决问题
我需要根据特定的catid列出所有vendorname.为此,我有2个表子类别和vendorSubCat.我试图通过首先根据catid获取subcatid(即id from subcategory table)来链接这两个表,然后根据从上一个表中选择的subcatid(来自vendorSubCat表)显示供应商名称.
表视图子类别
id subcatname subcatdesc catname catid
表视图vendorSubCat
vendorname vendorid subcatid
码
<?PHP
ob_start();
require_once('config.PHP');
$catid = $_REQUEST['catid'];
$sql1 = "SELECT * FROM subcategory where catid='" . $catid . "'";
$result1 = MysqL_query($sql1);
while ($row = MysqL_fetch_array($result1)) {
$myid = $row['id'];
if (sql1 != '0') {
$productt = MysqL_query("select * from vendorSubCat where id = '" . $myid . "' ");
$posts = array();
if (MysqL_num_rows($productt)) {
while ($post = MysqL_fetch_assoc($productt)) {
$posts[] = $post;
}
header('Content-type: application/json');
echo stripslashes(json_encode(array('list' => $posts)));
} else {
header('Content-type: application/json');
echo stripslashes(json_encode(array('list' => 'No productlist')));
}
}
}
?>
尽管代码工作正常,但结果显示在不同的数组中.它显示了这样的结果
{"list":[{"id":"1","vendorname":"Marzoogah","vendorid":"1","subcatid":"4"}]}{"list":[{"id":"2","vendorname":"Zee Zone","vendorid":"2","subcatid":"4"}]}{"list":[{"id":"3","vendorname":"Zee Zone","vendorid":"2","subcatid":"7"}]}{"list":[{"id":"4","vendorname":"????? ????????","vendorid":"3","subcatid":"4"}]}
我希望将所有结果放在一个数组中,即所有列表都应该放在一个列表中.每次显示新行时都不应显示“list”.任何帮助,将不胜感激
解决方法:
您无需立即回显结果:
echo stripslashes(json_encode(array('list' => $posts)));
而是将所有数据收集到一个数组:
$results = array();
//Your code
$results[] = array('list' => $posts);
//...
$results[] = array('list' => 'No product list');
//...
//And echo just one time in the end:
echo stripslashes(json_encode($results);
或类似的东西用于合并:
$results = array();
//Your code
$results = $results + $posts;
//...
$results = 'No product list';
//...
//And echo just one time in the end:
echo stripslashes(json_encode(array('list' => $results)));
就像是:
SELECT vsc.* FROM vendorSubCat vsc
INNER JOIN subcategory sc ON vsc.id=sc.id
WHERE sc.cat_id = 15