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I am facing some problem in my PHP function.
function pagename($id){
$query=mysql_query("select * from tbl_pages where recid='$id' and langid='$LangID'")
$rs=mysql_fetch_array($query);
$page_name=$rsp['pgname'];
print $page_name;
}
i am not getting any resutl
I saw some problems in your code
First, there is no connection string to your database, I hope you do the connection before to do the request.
Then, in your request, you try to use a variable $LangID not declared in your function, maybe you forgot to put it in your function declaration.
You put the result of your request in the $rs variable and then you try to read the $rsp variable.
You are using mysql in your code, actually it's very unsafe to use it, it's very recommended to use mysqli or PDO instead.
Finally, you don't return anything with your function, you are missing the return statement or maybe you just want to display the result ?
EDIT : I suggest you to write your SQL requests with uppercase, it's more readable for you and other people who read your code.
SELECT * FROM tbl_pages WHERE recid='$id' AND langid='$LangID'
There's no $LangID parameter.
$rs and $rsp are different variables in your code, you should use only one of them.
This function does not return anything, even if you get something in your print.
Check if mysql connection is already established.
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I'm PHP developer but i cant understand this error
$uid = $this->db->Tables("telegrambots")->search([
"telegrambotid" => $this->botKey
])['uniqueId'];
if (!file_exists("TelegramBotCommands/{$uid}"))
mkdir("TelegramBotCommands/{$uid}");
Eval is evil, you probably don't need it so don't use it. You want to make a call to a class with a dynamic name? Use this:
$dynamic_class_name = 'Video';
$video = new $dynamic_class_name();
That being said, your snippet with eval seems to work perfectly fine:
http://sandbox.onlinephpfunctions.com/code/e3bb43b1ccfd27365247120e9c5751aac9e2b4ce
You would have to check your logs as to what the error is.
EDIT:
As you said you are using namespaces, try to use the full classname including the namespace in the eval function (like new \namespace\Videos(.... Even better though: don't use eval!
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I am trying to pass some variables through url in php. Im using the following script
$query1 = new ParseQuery("Drivers");
$query1->includeKey("driverUserId");
$query1->descending("createdAt");
$results1 = $query1->find();
$object1 = $results1[$i];
$uname=$object1->get('driverUserId');
echo ''.$uname->get('name').'';
echo '<td>'.$uname->get('name').'</td>';
echo '<td>'.$uname->get('username').'</td>';
But when I execute the page, script stopped working in this line.
echo ''.$uname->get('name').'';
When I removed '.$uname.' it worked fine.
Because if $uname has a ->get() method, it's obviously a class or an object and you can't directly use it like a string (if it hasn't any tostring modifiers). Try to write it like this (just guessing):
echo ''.$uname->get('name').'';
Use the below code:
$id=$uname->get('id');
$name=$uname->get('name');
echo ''.$name.'';
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$qry = "SELECT `$speciality` FROM `graph`";
Could you change the code so that after executing the query you display the possible error
mysql_query($qry, $link);
echo mysql_errno($link) . ": " . mysql_error($link) . "\n";
this way you can see what goes wrong in the mysql code.
Also you're easily vulnerable to SQL injection attacks by inserting a variable in the query the way you are now.
Edit based on your comment that it is not the MySQL:
Also add error_reporting(-1); at the beginning of your php code to display any other errors. and use var_dump or print_r on suspicious variables to check what their value really is. This is the basics of debugging any php code.
you need to remove back-tick from your code around column name:-
$speciality = 'OBS';
echo $qry = "SELECT `".$speciality."` FROM graph";
Output:- SELECT `OBS` FROM graph
Note:- It strange because As #Jay says you code is working fine too, and now I check it too and it works fine. So I think may be something wrong in your next line of code. So check yourself. Thanks.
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I need to access the input value of a dynamic variable based on another input field.
Sample code:
$upid=$_GET['upid'];
$check_box_name='c'.$upid;
echo $upid;
$check=$_GET[$check_box_name];
any idea how do i access it??...Please help
The code you entered should work, but, it is vulnerable to errors, as you are dealing with a user input, you should either do a validation or a failover value.
If you are using PHP5.3+, you can easily do this as follows:
if ($check = #$_GET['c' . (#$_GET['upid'])]? : false !== false) {
//do something with $check
} else {
//failed
}
The # sign is used to omit any error or exception from being thrown. Also, it maybe a good thing to escape the $check variable for more security.
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I am getting this error
Notice: Undefined index: maxvalid in C:\wamp\www\myproj\includes\func.php on line 26
and my code is
require("common.php");
$incquery = "select max($TabFld) as maxvalid from $TabName";
$stmt = $db->prepare($incquery);
$incresult = $stmt->execute();
$row=$stmt->fetchAll();
$maxvalid = $row['maxvalid'];
if($maxvalid <> NULL)
{
$incvalid=$row['maxvalid']+1;
}
return $incvalid;
I am using PDO to connect mysql and I never used it before. I always use mysql_connect to connect database and I cannot understand why I am getting this error.
I also debug the code and see that value is not coming in $maxvalid variable but it came when I use mysql_connect.
fetchAll returns an array of rows. Try just fetch
For future reference, if in doubt, var_dump it.