MySQL 中的 PHP 更新查询未更新但 SELECT 和 INSERT 有效

Question

If I try to update an entry in my database, it doesn't work, as in not update. But when I try to select a value from the same database, that works just fine. I'm not sure why this is so.

Update query:

$id = 1;

try {
    $conn = db();
    $sql = "UPDATE instagram SET token=?, expires=? WHERE id=$id";
    $stmt = mysqli_stmt_init($conn);

    if (!mysqli_stmt_prepare($stmt, $sql)) {
        throw new Exception($conn->error);
    }

    mysqli_stmt_bind_param($stmt, "si", $tokenAccess, $tokenExpires);
    mysqli_stmt_execute($stmt);

    // on error
    if (!mysqli_stmt_execute($stmt)) {
        throw new Exception($conn->error);
    }

    var_dump($stmt);

    $stmt->close();
    $conn->close();
}
catch (Exception $e) {
    print_r($e);
}

var_dump for $stmt returns:

 object(mysqli_stmt)#4 (10) { ["affected_rows"]=> int(1) ["insert_id"]=> int(0) ["num_rows"]=> int(0) ["param_count"]=> int(2) ["field_count"]=> int(0) ["errno"]=> int(0) ["error"]=> string(0) "" ["error_list"]=> array(0) { } ["sqlstate"]=> string(5) "00000" ["id"]=> int(1) }

It's probably in plain sight and I'm missing the obvious. A pointer in the right direction is much appreciated.

UPDATE:

My table setup is as follows:

It has one entry where:

• token = ABCdefg123adc........ <-- example string, contains numbers and letters
• expires = 60
• created_at = 2019-12-01 08:57:26
• id = 1.

I tried to access the table via:

$sql = "INSERT INTO instagram (token, expires) VALUES (?, ?)";

This, like SELECT , works.

I also found that if (!mysqli_stmt_execute($stmt)) was sending the entry twice, so I took that out. Still, updating the entry doesn't work.

I have also echoed the values for $tokenAccess and $tokenExpires . The result contains 60 for $tokenExpires and the long-lived access token for $tokenAccess , as expected.

Answer 1

I was having similar problems when I was learning mysqli years ago. I am guessing that no one told you can enable error reporting for mysqli functions. You can do it by adding this line before you open a connection:

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);

I can see you are adding a lot of unnecessary code. You don't need all these if statements and you don't need to throw exceptions manually. PHP can do it for you.

You don't need to close the statement and you don't need to close the connection. Never catch exceptions just to print out the error message!

Using OOP is also going to make your code cleaner and easier to spot mistakes.

I can't see what your db() function looks like, but it should hopefully look something like this. Take a look how much simpler this code is:

function db(): \mysqli {
    mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
    $mysqli = new \mysqli('localhost', 'username', 'password', 'testdb');
    $mysqli->set_charset('utf8mb4'); // always set the charset
    return $mysqli;
}

$conn = db();

$id = 1;
$sql = "UPDATE instagram SET token=?, expires=? WHERE id=?";
$stmt = $conn->prepare($sql);
$stmt->bind_param('sss', $tokenAccess, $tokenExpires, $id);
$stmt->execute();

About your real problem:
There probably was no problem at all. affected_rows indicates that 1 row has been updated.

Answer 2

You are using prepared statements so, your codes in where clause should be like this WHERE id = ?";

$id = '1'; 
$tokenAccess = $_POST['token']; 
$tokenExpires = $_POST['expires'];
//Make sure you are passing these values to php 
try {
    $conn = db();
    $sql = "UPDATE instagram SET token=?, expires=? WHERE id = ?";
    $stmt = mysqli_stmt_init($conn);

    if (!mysqli_stmt_prepare($stmt, $sql)) {
        throw new Exception($conn->error);
    }

    mysqli_stmt_bind_param($stmt, "ssi", $tokenAccess, $tokenExpires, $id);
    //If $tokenExpires uses - OR / OR ETC.. you need to set it to varchar 
    // on error
    if (!mysqli_stmt_execute($stmt)) {
        throw new Exception($conn->error);
    }
    //you can return affected rows mysqli_stmt_affected_rows($stmt);
    var_dump($stmt);

    $stmt->close();
    $conn->close();
}
catch (Exception $e) {
    print_r($e);
}

See here : How to debug Php code?

Here is how you can use : https://www.php.net/manual/en/mysqli-stmt.affected-rows.php