收到警告:implode():收到 var 值后

问题描述

我遇到以下错误:

警告:array_map():参数#2 应该是一个数组...

警告:implode():在第 1918 行 wp-includes/class-wp-query.php 中传递的参数无效

Warning : implode(): Invalid arguments passed in wp-includes/class-wp-query.php on line 1918

在我的 php 上，我从另一个页面收到一个 var，我这样做了:

On my php I am receiving a var from another page and I do:

$myId = $_POST['varPostId'];
$parts = explode(',', $myId);

然后我需要读取该值

query_posts(array(
    'post__in' => $myId
));

但我收到了上述错误.

如果我这样做:

    'post__in' => $parts
));

我得到一个空白页.

我尝试使用 implode

$myId = $_POST['varPostId'];
$parts = implode(',', $myId);

并直接在查询中获取值

query_posts(array(
    'post__in' => $_POST['varPostId']
));

$_POST['varPostId'] 是一个单一的值，如 405

$_POST['varPostId'] is a single value like 405

推荐答案

您从 POST 请求中获得的值似乎只包含一个应该是 int 的帖子 ID，但是可能是一个字符串.您需要进行一些类型和错误检查.像下面这样的东西应该可以工作.我看到 query_posts 和 post__in，我假设 WordPress 在这里发挥作用?

It seems that the value you're getting from the POST request is containing just a single post ID which should be an int but is possibly a string. You'll need to do some type and error checking. Something like the following should hopefully work. I see query_posts and post__in, I'm assuming WordPress is at play here?

// Put your ID into a variable and sanitize it
$myId = sanitize_key( $_POST['varPostId'] );

// Force it to an int (in case it's not from sanitize_key)
$myId = (int) $myId;

// Don't explode because it's not appropriate here
// $parts = explode( ',', $myId );

// Add the post id to the arguments as an array
query_posts(
  array(
    'post__in' => array( $myId )
  )
);

为了跟进为什么 explode 在这里不合适，它用于转换字符串，通常是这样的:

To follow up on why explode isn't appropriate here, it's used to convert a string, usually something like this:

哦，你好，世界

放入一个包含多个项目的数组，如下所示:

Into an array with multiple items like the following:

$asdf = array(
  0 => 'oh',
  1 => 'hello',
  2 => 'world'
);

但是您似乎没有逗号分隔的帖子 ID 字符串，您只有一个，所以最好不要在这里使用.

But you it seems that you don't have a comma separated string of post ids, you just have the one so it's best not used here.

如评论中所述，explode 接受一个字符串并将其拆分为一组项目.implode 正好相反.它接受一个数组并将其压缩为一个字符串.

As already stated in the comments, explode takes a string and splits it into an array of items. implode does the opposite. It takes an array and condenses it down into a string.

所以你可以有这样的东西:

So you could have something such as:

$myArray = array(
  0 => 'hello',
  1 => 'world'
);

// Convert to a string
$myString = implode( ', ', $myArray );

echo $myString; // prints: hello, world

// Convert back to an array
$myString = explode( ', ', $myArray );

var_dump( $myString ); // prints the array above

这篇关于收到警告:implode():收到 var 值后的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！