为什么这个PHP数组只返回第一个记录？

Question

我试图从我的MySQL数据库中返回四条记录，但是只返回了第一条记录。我已经搜过了，但我不知道为什么。有人能给我指明正确的方向吗？

    <?php
session_start();
function displayImage($username){
    $imageDate  = $_POST['imageDate'];
    $result = mysql_query("
        SELECT
            imageName
        FROM
            images AS i
        INNER JOIN
            users AS u ON i.userID = u.UserID
        WHERE
            u.username = '$username'
        AND
            i.imageDate = '$imageDate'

    ") or die(mysql_error());

    //return a message to the users explaining ......
        if (!isset($_POST['Submit'])) {
            // this does nowt yet!!!
            $output = "Nothing selected yet.";
        }
        else {
        //This is a while loop to store the SQL results into ......
        $row = array(mysql_fetch_assoc($result));
        foreach ($row as $picName) {
        $cam = $_POST['cam'];
        $fullPath = $username . "/" . $cam . "/" . $picName['imageName'];
        // $output = //this works fine
                reset($images);
            }
        }
        var_dump($row);
        echo "<br />";
        return $output;
    }
?>

Answer 1

正如这里的其他人说的那样，您需要使用一个while循环来完成这个任务，我已经整理了一些代码，并添加了一些其他的内容供您考虑。

这样做的实际原因是，当您使用mysql_fetch_assoc时，它会返回一个结果资源，并从它留给返回的资源中删除它。所以，如果你试着把它存储在一个数组中，你就会得到数组中的第一个，而没有别的东西。当您使用while循环时，它的工作方式基本上是：“如果mysql_fetch_assoc有什么要提供的，那么在循环中执行代码”。

<?php

session_start();

function displayImage($username) {
    // Also, as others have said, check out PDO, or MySQLi, as they
    // both provide a better interface with MySQL an have better
    // security features.
    $username  = mysql_real_escape_string($username);
    $imageDate = mysql_real_escape_string($_POST['imageDate']);

    $result = mysql_query("
        SELECT
            imageName
        FROM
            images AS i
        INNER JOIN
            users AS u ON i.userID = u.UserID
        WHERE
            u.username = '$username'
    AND 
            i.imageDate = '$imageDate'
    ") or die(mysql_error());

    //return a message to the users explaining
    if (!isset($_POST['Submit'])) {
        // this does nowt yet!!!
        $output = "Nothing selected yet.";
    } else {
        $cam = $_POST['cam'];

        $images = array(); // This is part of the "you could do something 
                           // like" further down.
        while ($row = mysql_fetch_assoc($result)) {
            $fullPath = $username . '/' . $cam . '/' . $row['imageName'];
            // $output = //this works fine
            var_dump($row);
            echo "<br />";

            // Or you could do something like:

            $images[] = $username . '/' . $cam . '/' . $row['imageName'];

            // Then outside of this while loop you'd have all of your image 
            // paths stored in this $images array. 
            // 
            // It depends on how you want to handle outputting them.
        }
    }

    return $output;
}

代码中的注释贯穿我的主要观点。

我还将$cam =$_POST‘$cam’从循环内部移到了它的外部，因为它不需要进入循环，因为它的值总是相同的，不管您要遍历哪个循环项。

Answer 2

您应该使用循环来获取结果：

    while ($row = mysql_fetch_assoc($result)) {
        $cam = $_POST['cam'];
        $fullPath = $username . "/" . $cam . "/" . $row['imageName'];
    }

Answer 3

使用mysql_fetch_assoc或mysql_fetch_array的方法如下：

while ($row = mysql_fetch_assoc($result)) {
    // This is an example...
    echo $row["id"];
    echo $row["name"];
}