Question

如何在表单选项值中显示变量$ player-＆gt; name，以便用户可以选择变量并提交表单。

这是我的代码无效：

<?php
    $team = $_POST['team'];

    $result =        file_get_contents("http://nhlwc.cdnak.neulion.com/fs1/nhl/league/teamroster/$team/iphone/clubroster.j    son");
    $json = json_decode($result);
    $goalies = $json->goali;
    foreach ($json->goalie as $player) {

        **echo "<option value=\"".$player->name."\">".$player->name."</option>**
    }
?>

Answer 1

在声明结尾处没有双引号和分号。

 <?php 
$team = $_POST['team'];
$result =    file_get_contents("http://nhlwc.cdnak.neulion.com/fs1/nhl/league/teamroster/$team/iphone/clubroster.json");
$json = json_decode($result);
$goalies = $json->goali;
foreach ($json->goalie as $player) {

    echo "<option value=\"".$player->name."\">".$player->name."</option>";
}
?>

Answer 2

$json->goalie中的foreach值是什么，请改用<{1}}

$goalies

Answer 3

除了你的代码中有明显的解析错误之外，还有一些需要注意的事项：

$team = $_POST['team'];

$result = file_get_contents("http://nhlwc.cdnak.neulion.com/fs1/nhl/league/teamroster/$team/iphone/clubroster.json");

你不能在这样的网址中使用$team，你应该编码它：

$url = sprintf("http://nhlwc.cdnak.neulion.com/fs1/nhl/league/teamroster/%s/iphone/clubroster.json",
    urlencode($team)
);

$result = file_get_contents($url);

$json = json_decode($result);
foreach ($json->goalie as $player) {
    echo "<option value=\"".$player->name."\">".$player->name."</option>";
}

您应该始终逃避输出中的变量：

    printf('<option value="%s">%1$s</option>',
        htmlspecialchars($player->name, ENT_QUOTES, 'UTF-8')
    );

Answer 4

如果你想在字符串中显示变量，那么总是使用＆＃34;＆＃34;用于显示变量。您可以将代码修改为：

echo "<option value='{$player->name}'>{$player->name}</option>";

所以永远记住： 1.您可以使用＆＃39;（单曲）＆＃34;＆＃34;＆＃34;（双引号）

Answer 5

我建议将JSON解码为数组，而不是对象，它更容易使用。
您在上面的代码中没有使用goalies变量，我假设您要在foreach语句中使用它？
错误检查是个好主意

如果是这样，那么您的代码可能如下所示：

<?php
    $team = $_POST['team'];

    $result = file_get_contents("http://nhlwc.cdnak.neulion.com/fs1/nhl/league/teamroster/$team/iphone/clubroster.json");
    $json = json_decode($result, true);
    $goalies = array();
    if (!empty($json['goali'])) {
        $goalies = $json['goali'];
    }

    foreach ($goalies as $player) {
        echo '<option value="' . $player['name'] . '">' . $player['name'] . '</option>'
    }
?>

Answer 6

您没有正确使用'变量。

Demo

<?php

// Create a dummy Object
$player = new stdClass();

// Create a dummy property
$player->name = "Occam's Razor";

// Print it out
echo '<option value=" '.htmlspecialchars($player->name).' ">' .htmlspecialchars($player->name). '</option>';

这打印出以下内容。

<option value=" Occam's Razor ">Occam's Razor</option>

Answer 7

不是写下问题是什么，我认为向你展示更有帮助，并给你一些指示：

// Make sure all errors are displayed
error_reporting(E_ALL);
ini_set('display_errors', 1);

// Get the team variable and make sure that it was set
$team = filter_input(INPUT_POST, 'team', FILTER_UNSAFE_RAW);
if ($team === null) {
    die('Error: The "team" POST variable is not set, cannot continue');
}
// Since you are using the variable inside a URL it makes
// sense to remove potentially unsafe characters; or in
// other words, only preserve characters that are allowed.
// Encoding the variable, as in Ja͢ck's answer, is also an option.
$team = preg_replace('/[^a-zA-Z0-9_-]/', null, $team);
if ($team === "") {
    die('Error: The "team" variable is empty, cannot continue');
}
// Construct the URL
$url = "http://nhlwc.cdnak.neulion.com/fs1/nhl/league/teamroster/{$team}/iphone/clubroster.json";

// Grab the data from the website
$data = file_get_contents($url);
if ($data === false) {
    die('Error: Could not grab the data from the URL, cannot continue');
}
// Attempt to decode the data
$json = json_decode($data);
if (json_last_error() !== JSON_ERROR_NONE) {
    die('Error: There was a JSON error ('.json_last_error_msg().'), cannot continue');
}
// Since we are expecting an object we should check for it
// before using it
if ( ! is_object($json)) {
    die('Error: The JSON was decoded, but is not an object, cannot use it...');
}
// If the following code gives you problems then read the error
// message, try to understand what it says, and then google it if
// you are getting nowhere
$goalies = $json->goali;
foreach ($json->goalie as $player) {
    echo "<option value=\"".$player->name."\">".$player->name."</option>";
}

PHP数组：使用选项值在表单内回显$ var

7 个答案:

Demo