I have string like "Venditoris: Beware of Scams » Blog Archive » Trilegiant Complaints ..." in Database but when I try to display it ,It is not displaying.
So,I used html_entity_decode function but still it is not display.
I am Using cakePHP.below is my code to display that links.
echo $html->link(html_entity_decode(
$listing_end_arr[$i]['Listing']['listing_title'],ENT_QUOTES),
$listing_end_arr[$i]['Listing']['listing_url'],
array('target'=>'_blank', 'style'=>'color:'
. $colorArr[$listing_end_arr[$i]['Listing']['listing_sentiment']])) ;
Please Help me.
Check the CakePHP manual if you are using $html->link correctly. If so, var_dump the return value instead of echoing it. If it is empty, do
var_dump( $listing_end_arr[$i]['Listing'] );
to see what the Listing key contains. If the desired content is not in the dump, you know the error is elsewhere; probably in fetching the string from where it is stored.
Also, instead of using array[n][foo][bar][baz], consider assigning the subarray to a variable while looping over the array, e.g. $listing = array[n][foo][bar], so you can just do $listing[baz]. This will dramatically increase readability of your code.
inspect generated html first.. your code should echo a link, maybe it's just not visible (styling, color..).
Related
This is example what I mean:
I wanna grab result from this url web1.com/do.php?id=45944
Example output:
"pk":"bn564vc3b5yvct5byvc45bv","1b":129,"isvalid":true,"referrer":true,"mobile":true
Then, I will show data result on other site web2.com/show.php
But I just wanna see data value "pk", "1b" and "isvalid". I don't need "referrer" and "mobile" data.
So, when I access web2.com/show.php, it just show data like this:
bn564vc3b5yvct5byvc45bv 129 true
file_get_contents web1.com/do.php?id=45944
Grab this result "pk":"bn564vc3b5yvct5byvc45bv","1b":129,"isvalid":true,"referrer":true,"mobile":true
Filter and show value "pk", "1b" and "isvalid" on web2.com/show.php
So, can you help me with simple php code/script?
Sorry if you don't understand because my english.
Is this source providing JSON-formatted data? That is, with { }'s around it?
If so, use PHP's built in json_decode function. (Man page at http://php.net/manual/en/function.json-decode.php) It will parse the JSON data for you and return an associative array. For example
$your_JSON_data = '{"pk":"bn564vc3b5yvct5byvc45bv","1b":129,"isvalid":true,"referrer":true,"mobile":true}';
$your_array = json_decode($your_JSON_data);
echo $your_array['pk'] . ' ';
echo $your_array['1b'] . ' ';
echo $your_array['isvalid'];
If, for some reason, there are no JSON-style curly braces around your data, you can append them.... for example,
$proper_JSON = "{$bad_json_data}";
however, if that's the case, I'd wonder why it wasn't properly formatted in the first place. In that case it may be safer to use the PHP explode function (http://php.net/manual/en/function.explode.php)
i have a url.. eg http://www.example.com/index.php?rou=feed/web_api/categories&key=4&parent=0&level=10
when i hit the link simply on browser it displays a long list of multidimensional array.
I want to know that is it possible to store the url in a variable, something like this
$link=http://www.example.com/index.php?rou=feed/web_api/categories&key=4&parent=0&level=10
And then display the array in it as
echo '<pre>';
print_r($link);
echo '</pre>';
and then fetch the values from the array
I tried doing so but it displayed only a blank screen. Not sure if i am going on the right track. can anyone guide me
I think you are looking for one of these:
$data = file_get_contents('http://example.com/data.php');
This will fetch the raw data from the specified URL, and store the contents to the variable.
You may then need to implement your own code to actually parse the raw data, depending on your encoding type.
I'm trying to mix <?php echo do_shortcode('[...]') with a field from Advanced Custom Fields within Wordpress.
So basically what I'm trying to do is give the user a text field in the page edit screen where she can paste in the ID of a youtube vide. This field will then update my do_shortcode to display the correct video.
I'm not sure what I'm doing wrong considering I've done this several times before and been succesful. I do have a feeling I'm not escaping the string correctly?
<?php echo do_shortcode('[video_lightbox_youtube video_id="' . the_field("youtube_video") . '" width="640" height="480" anchor="Play Video"]'); ?>
Anyone able to lead me in the right direction? :)
EDIT
The code above returns q_cHD1lcEOo with multiple spaces in front of it as well as this: "Error! You must specify a value for the Video ID, Width, Height parameters to use this shortcode!" That's why I was thinking I'm not escaping it correctly as these are all specified.
I'll add that if I remove the_field("...") and replace it with just an ID it displays perfectly.
SECOND EDIT
Since I was not supposed to echo it, I was using the wrong function to get the field. Instead of using the_field() which prints the value, I was supposed to use get_field() to simply return it to the string.
Your question is somewhat unclear, but I'm also 20 hours without sleep.
Anyways, as far as mixing PHP within a PHP string, there's numerous ways to do it..
You can use concatenation or { } within the string itself.
For example, say we want to echo out the property of an object within a string.
We could do the following
echo "This is my property " . $object->property;
Or, we can do this
echo "This is my property {$object->property}";
You can even do cool things like access associative arrays within strings like so
echo "This is my property {$object->property['cool']}";
Hopefully this leads you in the ride direction.
At first glance it looks like you should be using get_field instead of the_field. the_field will print without being prompted, whereas get_field will return its value, which is what you want.
I see you've also mentioned whitespace at the start, you should consider wrapping the function in trim.
See below:
<?php echo do_shortcode('[video_lightbox_youtube video_id="' . trim(get_field("youtube_video")) . '" width="640" height="480" anchor="Play Video"]'); ?>
I am trying to convert a drupal installation into a front end driven by Code Igniter. This is an experimental project to check the performance boost I can get. But the biggest problem I am facing is that few fields in Drupal store php string as it is. For example
<?php print "A"; ?>
This is normal
Now I am able to see the text "This is normal" which comes from the query that I run in Code Igniter, but I don't see the php function which is saved inside the table. I can see the text when I view the record through phpmyadmin. But somehow not inside the CI query result.
I hope U've got the solution. But just in case if you haven't try this:
I created a table with two columns
//$result Contains the data fetched from the table using a model.
foreach ($result as $key=>$val) {
if($key == 'code') {
$val = str_replace('<?php','',$val); //Remove PHP's opening tag.
$val = str_replace('?>','',$val); //Remove PHP's closing tag.
$val = rtrim($val); //Remove leading and trailing spaces.
echo $key.': ';
eval($val.';'); //Execute the PHP code using eval.
} else {
echo $key.': '.$val.PHP_EOL;
}
}
I tried
echo $result['code']
print_r($result)
var_dump($result)
highlight_string($result['code'])
eval($result['code'])
and finally str_replace followed by eval($result['code']).
Check the screen shot of the result obtained from: here
There you can see that the Result produced by 1,2, and 5 are empty. But when you inspect element against the empty space It'll clearly show that the string that's echoed/print is commented out.
Screen-Shot. This is has nothing to do with codeigniter. Its done by HTML Parser.
So the solution is to remove the opening and closing tags of PHP and then use eval. I hope this helps.
Thanks for the help. Yes, the last resort was the eval function and that is the one which helped me attain which I was wanting it to do.
The data which was inside the database had PHP function and only eval function was able to treat that part as a PHP code and execute it when I was getting the data inside my view.
I'm trying to make a dynamic menu in my web, in which only some pages from each section will appear.
The code I wrote was:
$menulist=array();
$menulist[1]='file1%#16';
$menulist[2]='file2%#9';
$menulist[3]='file3%#19';
$menulist[4]='file4%#8';
$menulist[5]='file5%#13';
$menulist[6]='file6%#14';
$menulist[7]='file7%#10';
$menulist[8]='file8%#23';
$menulist[9]='file9%#19';
$menulist[10]='file10%#18';
$menulist[11]='file11%#12';
function actualizaciones($matriz)
{
$linea=explode("%#",$matriz);
echo '<li><a href="first_chunk_of_URL'.$linea[0].'middle_chunk_of_url'.$linea[1].'last_chunk_of_URL">'.${$linea[0]}[$linea[1]].'</li>;
}
echo '<ul>';
array_walk($menulist,'actualizaciones');
echo '</ul>';
Every $linea[0] string is the name of another array (not shown in this code) which contains the text that should be in every possible link corresponding to every key passed by $linea[1].
I must have done something wrong, because the hyperlinks work fine but there's no text showing on them.
use the simple character like below
echo '<li><a href="first_chunk_of_URL'.$linea[0].'middle_chunk_of_url'.$linea[1].'last_chunk_of_URL">'.${$linea[0]}[$linea[1]].'<li>';
and the problem in your code is
.'</li>;
^^^^^
here is the problem it should be
.'</li>';
If I'm reading the question right, you're asking how to use variable variables in PHP.
This can be done using the double-dollar syntax - ie $$linea[0]. See the PHP manual for more info: http://uk.php.net/manual/en/language.variables.variable.php
But if that is what you're doing, I would say you're not writing good code: if variable variables are involved, there's almost always a better way of doing it.
Can't really offer much better assistance here without understanding more about what you're trying to do, but it sounds like you should be using subarrays rather than separate named variables for everything.
Hope that helps.