【发布时间】:2015-04-21 07:33:50
【问题描述】:
这是我的代码,当我 ECHO 输出它的 show all POST array 时,但是当 mysqli_query 操作运行时它只插入最后一个值。如何在查询中插入所有数据?谁能帮帮我...
$ser = $_POST['serial'];
foreach ($ser as $seria) {
echo $serial = $seria;
}
$re = $_POST['ref_no'];
foreach ($re as $refe) {
echo $ref = $refe;
}
$des = $_POST['desc'];
foreach ($des as $desce) {
echo $desc = $desce;
}
$uni = $_POST['unitss'];
foreach ($uni as $units) {
echo $unit = $units;
}
$qt = $_POST['qty'];
foreach ($qt as $qtys) {
echo $qty = $qtys;
}
$pric = $_POST['price'];
foreach ($pric as $prices) {
echo $price = $prices;
}
$amoun = $_POST['amount'];
foreach ($amoun as $amounts) {
echo $amount = $amounts;
}
mysqli_query($con, "INSERT into purchase_order (po_id, po_no, serial_no, ref_no, description, unit, qty, price, amount, status) VALUES ('', '".$po_nom."', '".$mr_no."', '".$serial."', '".$ref."', '".$desc."', '".$unit."', '".$qty."', '".$price."', '".$amount."', 'Pending')");
【问题讨论】:
-
你的表单结构是什么?
-
这不是完成任务的正确方法...您可以打印所有内容,但只能插入一个内容,因为您尝试插入循环的最后一个值..