Question

剧本：

<?php

if(isset($_POST['submit'])) {

    $some_text_1 = $_POST['some_text_1'];
    $some_text_2 = $_POST['some_text_2'];
    $some_text_3 = $_POST['some_text_3'];

    $myarray = array($some_text_1, $some_text_2, $some_text_3);

    for ($i = 0; $i < count($myarray); $i++) {

        $tqs = "INSERT INTO `table` (`some_text`) VALUES ('" . $myarray[$i] . "')";
        $tqr = mysqli_query($dbc, $tqs) or die(mysqli_error($dbc));


        $tqs = "SELECT `id` FROM `table`  WHERE `some_text` = '" . $myarray[$i] . "'";
        $tqr = mysqli_query($dbc, $tqs) or die(mysqli_error($dbc));
        $fetch_array[] = mysqli_fetch_array($tqr);

    }

    $fetch_array = implode(", ", $fetch_array);

?>

我希望将数组插入到一行中，如下例所示：

the_row: | 3, 4, 5, 6, 7

虽然在使用implode函数时我收到此错误：

Notice: Array to string conversion in ... (points to the implode function)

有关如何使其正常工作的任何建议吗？

Answer 1

implode()需要一个字符串数组作为第二个参数。你给它一个数组数组。

你可以用这样的东西来解决这个问题（只是一个例子）：

$row = mysqli_fetch_array($tqr);
$fetch_array[] = row['id'];

错误：＆＃34;数组到字符串转换＆＃34; - 内爆功能

1 个答案: