How to access foreach loop variable in php?

Question

There is the following code that works correctly, that is, the title of the post is displayed and after the comma its category name on the page in the header:

 <?php foreach (HTML_SOBI::getMyCategories($mySobi) as $category) { echo $category['name'];}?></h1>

How can I display a post with the same title, category ['name'] format in OpenGraph?

The following working code but no category name:

<meta property="og:title" content="'.$mySobi->title.'" />

How can I correctly add the category name after the title, separated by commas? This part of code: <?php foreach (HTML_SOBI::getMyCategories($mySobi) as $category) { echo $category['name'];}?>

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il4mb · Answer 1 · 2021-06-15T22:40:50.080

0

Just try this

 echo $mySobi->title.' , '.$category;

Apply it like this

$YouTitle = "MY TITLE";
$category = "MY CATEGORY";
$Meta = $YouTitle." , ".$category;

echo'<meta property="og:title" content="'.$Meta.'" />';

Or from an array

$source = array(
    "Title" => "My Title",
    "category" => "My Category"
    );

$meta = $source['Title']." , ".$source['category'];

echo'<meta property="og:title" content="'.$meta.'" />';

edited Jun 15 '21 at 22:40

answered Jun 15 '21 at 15:47

il4mb

105
1
5

The following working code but no category name: `` where to insert your solution? – Igor Jun 15 '21 at 20:59
How can I understand if I don't know where your category is from and what source your title is from?, you have to enter a script of where the category comes from – il4mb Jun 15 '21 at 22:44
`Apply it like this` works super! however, how to call the category name? `$category = "MY CATEGORY"; ` `$category = ""; ` ?? – Igor Jun 16 '21 at 07:33

shaedrich · Answer 2 · 2021-06-16T09:06:44.447

0

If I understood the question properly, this might be what you want:

$categories = array_map(function($category) {
    return $category['name'];
}, HTML_SOBI::getMyCategories($mySobi));
$opengraph_meta=' <meta property="og:title" content="'.$mySobi->title.' '.implode(', ', $categories).'" />';

edited Jun 16 '21 at 09:06

answered Jun 16 '21 at 07:58

shaedrich

5,457
3
26
42

syntax error, unexpected `<` – Igor Jun 16 '21 at 08:05
Oh, sorry. I wrapped it in an echo – shaedrich Jun 16 '21 at 08:07
My complete code : `$opengraph_meta=' ';` – Igor Jun 16 '21 at 08:11
Okay, I added your code to the answer – shaedrich Jun 16 '21 at 08:13
Why, when using this when further on the page, `` there is no output of the category name, but the value is shown instead `ARRAY`>? – Igor Jun 16 '21 at 08:29
Because it is an associative array. It has multiple categories with probably more fields than just the `name` in each category. And since an array is a [compound type](https://www.php.net/manual/en/language.types.intro.php), it cannot be directly `echo`d as string. – shaedrich Jun 16 '21 at 08:34
Where should you put a space between the name and the category? `` – Igor Jun 16 '21 at 08:45
before `implode` like this: `` – shaedrich Jun 16 '21 at 09:06
Are there any medots for referring to $categories other than the `implode`? To use `$categories` further down the page instead of `` ? – Igor Jun 16 '21 at 09:20
Depends on what you want to do. If you want to output this array as a string, there has to be some sort of "stringification" so I concatinated them with a comma. If you want to use them without outputting them, using $categories is of course fine. – shaedrich Jun 16 '21 at 09:22
Maybe there is a more beautiful solution than mine `
title; ?>,
` , considering that I now have a variable `$categories` – Igor Jun 16 '21 at 09:28
Yes, mine. Just use the same code that is used for the meta tag. – shaedrich Jun 16 '21 at 09:28
I'm sorry, but how to do it correctly? – Igor Jun 16 '21 at 09:38
`
title; ?>,
` – shaedrich Jun 16 '21 at 09:39

How to access foreach loop variable in php?

2 Answers2

title; ?>,

title; ?>,