Question

我正在尝试使用以下代码从我的数据库中创建一个变量（$ order_description）27行：

$sql_query1 = "SELECT order_description FROM single_user_orders WHERE username = '". $_SESSION['login_user'] ."'";
$result1 = mysqli_query($dbconfig, $sql_query1);
$row1 = mysqli_fetch_array($result1, MYSQLI_ASSOC);
$count1 = mysqli_num_rows($result1);

if($count1 >= 1) {
    while ($row1 = $result1->fetch_assoc()) {
        $order_description = $row1['order_description'];
    }
}

虽然这段代码正常，但是当我<?php echo $order_description; ?>它返回SELECT语句的最后一行而不是我应该看到的27行时，我哪里出错？

Answer 1

要查看所有27个描述，您需要在循环中打印它们或将它们添加到数组中。当您运行循环并仅在循环结束后输出单个变量时，它将显示分配的最后一个值。

while ($row1 = $result1->fetch_assoc()) {
    $order_description = $row1['order_description'];
    echo $order_description."\n";
}

OR

while ($row1 = $result1->fetch_assoc()) {
    $descriptions[]=$row1['order_description'];
}
print_r($descriptions);

注意

@RiggsFolly正确地指出你有一个额外的fetch_array调用被浪费，因此你永远不会看到第一行。摆脱它。

踢你代码的第三行

$row1 = mysqli_fetch_array($result1, MYSQLI_ASSOC);

Answer 2

使用以下方式使用group_concat并阅读注释

$sql_query1 = "SELECT group_concat(order_description separator ',') as order_description FROM single_user_orders WHERE username = '". $_SESSION['login_user'] ."'" ; // update thie query with group_concat
$result1 = mysqli_query($dbconfig, $sql_query1);
$count1 = mysqli_num_rows($result1);
$order_description = array();
if($count1 >= 1) {
    while ($row1 = $result1->fetch_assoc()) {
        $order_description = explode(',',$row1['order_description']); // explode the string 
    }

}

print_r($order_description);

使用MySQLi返回多行

2 个答案: