CODEWITHSUNDEEP
phpsql
user10703810
user10703810

Reputation:

Select all entries where a given string is a substring of a spesific column in MySQL/PHP

I want to query my SQL database for a name. Let's say a certain table contains the column "names" and has two inputs "John Silva" and "Dave Silva". I want to make a query where I get all inputs with the name "Silva" and store them in a way that I can later echo out the result onto my HTML code. Here's what I got so far (keep in mind that it is not working, that's why i came here to ask :) ):

    $query = "SELECT ID, email, fullname, permission FROM users WHERE name LIKE '$data'";
$result = mysqli_query($conn, $query);

        if(mysqli_num_rows($result) > 0){

                while($row = mysqli_fetch_assoc($result)){

                    $userName = $row['fullname'];
                    $userHashedPermission = $row['permission'];
                    $ID = $row['ID'];
                    $userEmail = $row['email'];
                }

                if(password_verify('sim', $userHashedPermission)){
                    $userPermission='sim';
                }else{
                    $userPermission='nao';
                }

                $callbackObj->name = $userName;
                $callbackObj->permission = $userPermission;
                $callbackObj->Id = $ID;
                $callbackObj->email = $userEmail;

                $callbackJson = json_encode($callbackObj);
                echo $callbackJson;
                
        }else{
echo "something went wrong"
}

Upvotes: 1

Views: 181

Answers (3)

Kunal Raut
Kunal Raut

Reputation: 2584

You need to make some changes in your query as

$query = "SELECT ID, email, fullname, permission FROM users WHERE name LIKE '%".$data."%'";

As per the comments

Do not use while loop keep your code as

 if(mysqli_num_rows($result) > 0){                  
    $row = mysqli_fetch_assoc($result));                                
    $callbackJson = json_encode($row);              
    echo $callbackJson;             
 }

Upvotes: 0

Joseph Godwin
Joseph Godwin

Reputation: 1

I think your sample code and the introduction are totally explaining different thing.

From your introduction, I think you are trying to get all search results with the same surname and then save them in another different way.

If that is your point, we will have better way to sort it out.

Upvotes: 0

Gordon Linoff
Gordon Linoff

Reputation: 1269493

The functionality you want is like. More importantly: learn to use parameters.

$query = "SELECT ID, email, fullname, permission FROM users WHERE name LIKE CONCAT('%', ?)";
$stmt = mysqli_prepare($conn, $query);
mysqli_stmt_bind_param($stmt, "s", $data)
$result = mysqli_stmt_execute($conn, $query);

Upvotes: 1

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