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"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 6 years ago.
I have a list of profiles, and i need to make the user able to select the individual profile in a link.
I found this question and tried to follow it, but it doesnt quite work as i hoped it would: Displaying a user profile page PHP
The file to display all profiles on the page looks like this:
foreach ($rows as $row) {
echo '<a class="viewProfile" href="user.php?id=' . $r['UID']. '"><button>View Profile</button></a>';
}
But when i click the link the UID doesn't get posted in the URL, and it just looks like this: domain/user.php?id=
Update: With help from you guys i got the UID posted into the URL. Thanks!
And as jothi stated $id=$_GET['id'];
That's because $r['UID'] is not defined. You are enumerating some $row, so I assume that you mean $row['UID'].
If it doesn't work, please, post the result of this line, just before you foreach ($rows as $row)
echo print_r($rows);
Related
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed last year.
I want to extract the information on this page using a script PHP Simple HTML DOM,
Here is part of the code I wrote:
$html = file_get_html('https://www.digikala.com/product/dkp-7475119/');
foreach ($html->find('p.color-900') as $e) {
$color = $e->outertext;
echo $color ;
break;
}
But unfortunately the output I receive is Undefined variable
As #Kevin Y said in his comment, you are searching for something that isn't there... yet.
Visit view-source:https://www.digikala.com/product/dkp-7475119/ and search for <p>
The page hasn't loaded the extra elements onto it yet like the <p> tag you are looking for. This is likely because the page is loading them in after-the-fact using JavaScript so you will need to find a way to get the html after the page has loaded its contents that are added in later.
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 2 years ago.
I'm having a problem with an undefined index error message in PHP. I have a form with several <select> inputs, and all of them work except for one. On one I get the undefined index error message. I don't understand why this is happening, as it is formatted exactly the same as the other <select> inputs. I've made sure that the $_POST assignment in the attached script is referencing the correct element name, and that the option values are correct in the HTML.
The code for the entire form is here: https://pastebin.com/52xtECCT
Here is the code for the <select> element:
<td>Team to Represent:</td>
<td><select name="txtTeamType" id="txtTeamType">
<?php
$resultSet = $conn->query("SELECT intTypeofTeamID, strTypeofTeam FROM TTypeofTeams");
while ($rows = $resultSet->fetch_assoc())
{
$intTypeofTeamID = $rows['intTypeofTeamID'];
$strTypeofTeam = $rows['strTypeofTeam'];
echo "<option value='$intTypeofTeamID'>$strTypeofTeam</option>";
}
?>
</select>
</td>
</tr>
And here is the code for the variable assignment. This is the code that throws the error: $intTypeofTeamID = $_POST["txtTeamType"];
Thanks for any help, I really appreciate it.
I found the solution. Changing $intTypeofTeamID = $_POST["txtTeamType"]; to
if (isset($_POST["txtTeamType"])) { $intTypeofTeamID = $_POST["txtTeamType"]; }
fixed the issue.
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 4 years ago.
I'm very new at php. I get these errors when I write these codes.
check to see if islem was passed in before trying to access it
if (isset($_POST['islem']) {
// your code here
}
Otherwise you won't be able to load the page up initially in order to submit the form. Also notice that I used _POST here, this is because you used the POST method in your form.
You are using the method POST in the form and getting the variables with GET
Either Try method : GET or try $_POST["islem"];
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 6 years ago.
I am aware that I can use sessions and GET/ POST method but I would like to achieve this using cookies. My code on page1.php is :
$_COOKIE['varname'] = $id;
and on page2.php is:
$id = $_COOKIE['varname'];
I get the following notice on my browser: Undefined index $id
What is the problem with my code?
Try using setcookie('varname', $id) then
if (isset($_COOKIE['varname']){ echo $_COOKIE['varname']; }
To set a cookie you need to use setcookie(). And it has to be done prior to any output.
setcookie("mycookie", "myvalue" , $validtime); // validtime is a integer.
This question already has answers here:
"Notice: Undefined variable", "Notice: Undefined index", "Warning: Undefined array key", and "Notice: Undefined offset" using PHP
(29 answers)
Closed 4 years ago.
This error I am getting in Laravel. Can I know what exactly can be the issue. I have spent my full day into this.
you should pass all your variables you use in view, if you use $title as a title for each page in your website you should pass it from your controller function to the view like this:
return View::make('home.index')->with('title', 'your title here');
and in your view you can make a condition to display public title to all pages don't have a title like this:
#if(isset($title)){{$title}} #else {{'Default title here'}} #endif
Check your Controller. If in controller variable $title is empty and you manipulate with it then you will get similar error.
your $title variable is null .
Make sure that $title is not null
try this in your blade
#if(isset($title)){{$title}}#endif