我正在尝试从数据库中选择多个产品并将 id 插入到另一个表中。
我尝试如下执行,如果我只是 echo $name,$_POST["txtMaterial"][$key]并且$_POST["txtSize"][$key]在查询之前,我会获取所有选定的名称、材料和尺寸,但是当我绑定查询中的值时,我只得到第一个。
foreach($_POST["txtName"] as $key => $name){
$sQuery = $db->prepare('SELECT id FROM products WHERE name = :sName AND material = :sMaterial AND size = :sSize');
$sQuery->bindValue(':sName', $name);
$sQuery->bindValue(':sMaterial', $_POST["txtMaterial"][$key]);
$sQuery->bindValue(':sSize', $_POST["txtSize"][$key]);
$sQuery->execute();
$aOrders = $sQuery->fetchAll();
foreach($aOrders as $aOrder){
echo print_r($aOrders);
}
}
的 print_r$aOrders是这样的:
"Array
(
[id] => 174
)
1"
这是我需要的第一个 id。谁能帮我看看如何获取所有的 id?
$_POST["txtName"]foreach之前的var_dump :
"array(50) {
[0]=>
string(0) ""
[1]=>
string(6) "BACURI"
[2]=>
string(0) ""
[3]=>
string(0) ""
[4]=>
string(0) ""
[5]=>
string(0) ""
[6]=>
string(0) ""
[7]=>
string(0) ""
[8]=>
string(0) ""
[9]=>
string(0) ""
[10]=>
string(0) ""
[11]=>
string(0) ""
[12]=>
string(0) ""
[13]=>
string(0) ""
[14]=>
string(0) ""
[15]=>
string(0) ""
[16]=>
string(0) ""
[17]=>
string(0) ""
[18]=>
string(0) ""
[19]=>
string(0) ""
[20]=>
string(0) ""
[21]=>
string(0) ""
[22]=>
string(0) ""
[23]=>
string(0) ""
[24]=>
string(0) ""
[25]=>
string(0) ""
[26]=>
string(0) ""
[27]=>
string(0) ""
[28]=>
string(0) ""
[29]=>
string(6) "CAJARI"
[30]=>
string(0) ""
[31]=>
string(0) ""
[32]=>
string(0) ""
[33]=>
string(0) ""
[34]=>
string(0) ""
[35]=>
string(0) ""
[36]=>
string(0) ""
[37]=>
string(0) ""
[38]=>
string(0) ""
[39]=>
string(0) ""
[40]=>
string(0) ""
[41]=>
string(0) ""
[42]=>
string(0) ""
[43]=>
string(0) ""
[44]=>
string(0) ""
[45]=>
string(0) ""
[46]=>
string(0) ""
[47]=>
string(0) ""
[48]=>
string(0) ""
[49]=>
string(0) ""
}
"
foreach之前的var_dump $_POST["txtMaterial"]:
"array(2) {
[0]=>
string(1) "3"
[1]=>
string(1) "2"
}
蛊毒传说