Question

我想根据mySQL数据库创建一个选择列表，但是当使用mysqli_fetch_array时我得到空结果。选项编号是正确的，但都是空的。

以下是代码：

$rows = mysqli_num_rows($result);
$self = $_SERVER['PHP_SELF'];
    <div id='editUser'>
        <span class='ACtitle'>Edit Users</span>
        <div id='addField'>
            <form action="<?php $self ?>" method="post">
                <select name="target" size="1">
                    <option value='none'>Select user you wish to edit.</option>
<?php
    for($ii = 0; $ii <= $rows; $ii++)
    {
        $sqlSta3 = "select name from 'users'";
        $result2 = mysqli_query($database, $sqlSta3);

        $nameArray = mysqli_fetch_array($result2);

        print_r($nameArray);

        $tableName = $nameArray['name'];
        print "<option value='$tableName'> $tableName </option>";

    }
?>

                </select>
                <input type="text" name="updateName"
            </form>
        </div>
    </div>

Answer 1

您需要将mysql_fetch_array替换为mysql_fetch_assoc才能使用您的方法。 mysql_fetch_array返回类似[0] => "data",[1] => "data"的数组 mysql_fetch_assoc返回包含表['username'] => 'data',['fname'] => "data";

中单元格名称的数组

这是固定代码：

$rows = mysqli_num_rows($result);
$self = $_SERVER['PHP_SELF'];
    <div id='editUser'>
        <span class='ACtitle'>Edit Users</span>
        <div id='addField'>
            <form action="<?php $self ?>" method="post">
                <select name="target" size="1">
                    <option value='none'>Select user you wish to edit.</option>
<?php
    for($ii = 0; $ii <= $rows; $ii++)
    {
        $sqlSta3 = "SELECT name From users ";
        $result2 = mysqli_query($database, $sqlSta3);

        $nameArray = mysqli_fetch_assoc($result2);

        print_r($nameArray);

        $tableName = $nameArray['name'];
        print "<option value='$tableName'> $tableName </option>";

    }
?>

                </select>
                <input type="text" name="updateName"
            </form>
        </div>
    </div>

mysqli_fetch_array返回空结果

1 个答案: