While retrieving values from array without index in PHP I am getting nothing.
From my database value is stored like ["SCHOOL"] and I want to get just SCHOOL from this.
It's not treated as array instead it's treated as a string.
What will be the solution if want to treat this as array and get value of that array.
My Code is as follows :
$stream_arr = $res_tbl['streams'];
which gives result as ["SCHOOL"]
and I want just SCHOOL from this.
I am using code as $stream = $stream_arr[0];
and I get '[' this as result.
If I assign manual value to $stream_arr = ["SCHOOL"],
above code works and returns SCHOOL as expected.
But for results from my database is not working where result is same as the one I manually assigned.
To eliminate braces from the string you can use below code.
$stream_arr = '["SCHOOL"]';
$temp=explode('"',$stream_arr);
$val=$temp[1];
echo $val; //it will dispaly 'SCHOOL'.
But check the insert query(from the place you store data into db) that why it is storing with braces? if it is done by you as required then proceed else first fix that issue instead of eliminating braces in php code.
Related
I am making a dashboard using php and displaying it on webpage using html. For this I am taking some count values. Is !$value[" variable"] valid in php? How do i get the not of (!=) value? There is no other way to get the data. I am getting the values from a database.
print "<td><a href=\"$strBase and status!='Closed'\" target=\"_blank\">".$value[" ? "]."</td>";
I need to print !closed. Somehow I do not know how to pass that inside $value[].
If you want to use a variable as a key in order to retrieve a value from an array you can do as follows (I use a snippet of your code):
.$value[$myVar].
Where $myVar should be a string with the name of the key you want to retreive from $value
But anyhow, your question is pretty unclear so I'm not sure if this is what you are looking for.
You can check if the variable is empty(), i.e.:
if(empty($value["variable"])){
echo "variable is empty";
}
Here's the link i want to decode:
https://api.instagram.com/v1/users/28855276/media/recent/?client_id=775829274b6f48c1ac2bf218dda60e16
Actually i've tried many methods to get the result i want but i failed to do so with PHP.
i need to extract a value from this json which should equal a certain variable
Example:
$variable = json_value
The Hard thing here that i want a duplicated value.
the value i need is:
profile_picture":"http:\/\/images.ak.instagram.com\/profiles\/profile_28855276_75sq_1348344197.jpg"
Do not know how to reach this value.
Here's it's place in the json file
{"username":"zedd","profile_picture":"http:\/\/images.ak.instagram.com\/profiles\/profile_28855276_75sq_1348344197.jpg","id":"28855276","full_name":"Zedd"}
Please note that i've the username and id values already known.
Need to extract the profile_picture link to a PHP variable.
Please note: profile_picture is duplicated.
There is no duplication by me.
<?php
$array = json_decode('{"username":"zedd","profile_picture":"http:\/\/images.ak.instagram.com\/profiles\/profile_28855276_75sq_1348344197.jpg","id":"28855276","full_name":"Zedd"}',true);
echo $array['profile_picture'];
?>
In my php script,i am using a php variable inside an sql query.The php variable acquires its value through a post variable as follows:
$desc0=$_POST['desc0'];
$desc1=$_POST['desc1'];
$desc2=$_POST['desc2'];
$desc3=$_POST['desc3'];
$desc4=$_POST['desc4'];
$desc5=$_POST['desc5'];
$desc6=$_POST['desc6'];
$desc7=$_POST['desc7'];
$desc8=$_POST['desc8'];
$desc9=$_POST['desc9'];
The query is:
for($i=0;$i<10;$i++)
{
$q="insert into photos(name,category,description) values ('{$name{$i}}','$category','{$desc{$i}}')";
}
The problem is that on submitting the form i am getting an error which says
"undefined variable desc".
Therefore its not taking the values from the previously defined variables?
Any help?
First of, you code is completely unsafe - you should not pass user data directly into your query. There are many topics about it, and this is a good start.
Next, you don't need to store your data in such weird way. What if you'll want to pass 20 photos? In HTML, name your fields like photos[] - and in PHP, your values will be correctly parsed as an array $_POST['photos'], so you will be able to work with array:
$photos = $_POST['photos'];
foreach($photos as $photo)
{
//$photo contains certain item, so handle it with your logic
}
Finally, your issue is because of non-obvious PHP possibility for array-dereference with curly brackets. So your $desc{$i} is an attempt to access $i-th index for non-existent array $desc. Either use $desc$i or use concatenation to separate your variables.
You must change $desc{$i} to ${"desc" . $i}
Look at the following code:
foreach($_SESSION['cart'] as $r => $stringrow) {
$rowarray=explode("#",$stringrow);
}
$_SESSION['cart'] contains the row of concatenated values the separator is "#". Which I got from the previous page (as you can see they are inserted in a session.) I am exploding $stringrow to get an array called $rowarray. Everything is fine till here. The problem is this $rowarray[] contains two array elements, each of which contain concatenated imageurls separated by "#". The concatenated image urls are in $rowarray[11] and $rowarray[12]. I am again exploding $rowarray[11] and $rowarray[12]. trying to access each imageurl. But it's not happening. till now I have used innumerable echos to see the content of the imageurl array but every time it's showing 'Array'. as the value.
To view the contents of arrays, try using either var_dump($array) or print_r($array). They'll show you the type and contents of whatever variable you feed to them. It makes debugging oddly nested arrays much easier.
Using PHP I'm populating a dropdown list with values taken from a MySQL database. The list displays fine, my problem comes when I try to retrieve the selected value. I'm defining a variable and passing it the dropdownlist name for the POST array:
$variable = $_POST['dropdownlist'];
but the contents of $variable are \{value}"
Why is it putting in the \{ and }", and how do I get rid of them to get a value I can actually use?
If you are getting curly brackets as part of the value then the chances are they are their in your HTML code. Your code is likely something like echo '{$value}'; rather than echo "{$value}"; or something similar.