₦airaland Forum

Pls Guys

I ave hit a stumbling block and i need ur help.

Here Is D Query

$res = "SELECT * FROM `user` WHERE `userid` = '$userid'"

$result = mysqli_query($conn,$res);


Now D Problem Is If i loop through $result to get d row result(d user information),
I Get Notin.

And NB: 1. $userid Is Gotten Thru Global Variable $Post[] Where User Input Thier Id.


2. If I Use Let Say '12' In Place Of $userid It Return D Info.

ANTONINEUTRON:
Pls Guys

I ave hit a stumbling block and i need ur help.

Here Is D Query

$res = "SELECT * FROM `user` WHERE `userid` = '$userid'"

$result = mysqli_query($conn,$res);


Now D Problem Is If i loop through $result to get d row result(d user information),
I Get Notin.

And NB: 1. $userid Is Gotten Thru Global Variable $Post[] Where User Input Thier Id.


2. If I Use Let Say '12' In Place Of $userid It Return D Info.

i dont Really understand your question but i think this will solve your problem
$result = mysqli_fetch_array($res);
$nameyourvariable = $result["yourrow"];

Dangujba1111:
i dont Really understand your question but i think this will solve your problem
$result = mysqli_fetch_array($res);
$nameyourvariable = $result["yourrow"];

It Is Not Displaying D Query(converting It To And Array) That Is D Problem, D Problem Is D Query Isn't Returning Any Result/Info That Will Be Display

ANTONINEUTRON:
It Is Not Displaying D Query(converting It To And Array) That Is D Problem, D Problem Is D Query Isn't Returning Any Result/Info That Will Be Display

firstly, you have check if the id exists or not using $check = mysqli_num_rows(mysqli_query($conn, "SELECT * FROM user WHERE userid = '$userid'");
if($check < 1){
echo "No ID exists";
}

ANTONINEUTRON:
Pls Guys

I ave hit a stumbling block and i need ur help.

Here Is D Query

$res = "SELECT * FROM `user` WHERE `userid` = '$userid'"

$result = mysqli_query($conn,$res);


Now D Problem Is If i loop through $result to get d row result(d user information),
I Get Notin.

And NB: 1. $userid Is Gotten Thru Global Variable $Post[] Where User Input Thier Id.


2. If I Use Let Say '12' In Place Of $userid It Return D Info.

1. Why are you putting the database table and table field userid in quote?
They would be seen as strings that way.
2. Your $userid variable shouldn't have quote , it represents integer values.

So i think your query should look like this:

$res="SELECT * FROM user WHERE userid=$userid"

noordean:
1. Why are you putting the database table and table field userid in quote? They would be seen as strings that way. 2. Your $userid variable shouldn't have quote , it represents integer values.
So i think your query should look like this:
$res="SELECT * FROM user WHERE userid=$userid"

Thanks I'll Try It Right Away

Please let me know if you are still having any problems.