这是我的桌子。单击“完成”时,它会选中已选中的复选框ID,并在文本框中显示。在PDF按钮上,单击它将文本框输入发布到另一页:
所以我想获取所有ID的详细信息,但是我的代码仅显示第一个ID的结果
if (isset($_POST['allids'])) {
$query = "SELECT * FROM tbl_order
WHERE order_id = '".$_POST['allids']."'";
$result = mysqli_query($conn, $query);
while($row = mysqli_fetch_array($result))
{
echo $row['order_no']."</br>";
echo $row['ordercustomer_name']."</br>";
}
}
这是我的表格
<form method="POST" action="fetch3.php" target="_blank">
<div class="container" id="order_data">
</div>
<div>
<h4><strong>GENERATE REPORT</strong></h4>
<input type="text" name="allids" id="allids">
<button type="button" class="btn btn-primary btn-lg btn_done" name="btn_done" id="btn_done"><i class='fa'></i> DONE</button>
<button type="submit" class="btn btn-warning btn-lg btn_pdf" name="btn_pdf" id="btn_pdf"><i class='fa fa-file-pdf-o'></i> PDF</button>
</div>
</form>
答案 0 :(得分:2)
您将在SQL查询中使用Set<Item> itemSet = new HashSet<>();
Map<Item,Integer> occurencesMap = new HashMap<>();
itemSet.add(yourItem);
itemSet.add(yourItem);
itemSet.add(secondItem);
//itemSet now contains only 2 objects!
if (occurencesMap.get(yourItem) == null){
occurencesMap.put(yourItem,1);
}
else{
//adding 1 to the current value of occurences for that item.
occurencesMap.get(yourItem)++;
}
语句,并用逗号分隔ID组:
IN
警告:Little Bobby说 your script is at risk for SQL Injection Attacks. 。甚至escaping the string都不安全!
了解有关prepared的MySQLi语句。专为WHERE order_id IN (695,392)
使用these instructions
答案 1 :(得分:-2)
您可以使用:
if (isset($_POST['allids'])) {
$query = "SELECT * FROM tbl_order
WHERE order_id in('".$_POST['allids']."')";
$result = mysqli_query($conn, $query);
while($row = mysqli_fetch_array($result))
{
echo $row['order_no']."</br>";
echo $row['ordercustomer_name']."</br>";
}
}
答案 2 :(得分:-2)
现在一切正常,谢谢大家
$id = $_POST['allids'];
$query = "SELECT * FROM tbl_order
WHERE order_id IN($id)";
$result = mysqli_query($conn, $query);
while($row = mysqli_fetch_array($result))
{
echo $row['order_id']."</br>";
echo $row['order_no']."</br>";
echo $row['ordercustomer_name']."</br>";
}