Pick a value from table and put in a variable

Question

I have a Table named 'progress' and inside a column 'number' with value '2'. I want to pick the value 2 and put in a $result.

$letter ='A';
$number = "SELECT number FROM progress";
$result = mysql_fetch_object($number);
$finalcode = $letter . $result;

plus, seeing a previous question of yours https://stackoverflow.com/q/47528452/1415724 you were using the mysqli api and using mysql_ here, why? you're not mixing here are you? — Funk Forty Niner, Dec 30 '17 at 16:20

ndrwnaguib · Answer 1 · 2017-12-30T16:41:51.533

0

You did not execute your query. So mysql_fetch_object() has nothing to return. Moreover, You need to connect to your database which is accomplished using mysql_connect(). To execute queries using mysql you can use mysql_query. Wrapping the previous notes in your code as an example would be:

// Connecting to database
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully'; // Connected

$letter ='A';
$number = "SELECT number FROM progress";

$result = mysql_query($number); // Executing your query
$row = mysql_fetch_object($result); // Fetching your result into 'row'

$finalcode = $letter . $row;

In the above code, your object was fetched into $row. You can also specify the class name as the second parameter in mysql_fetch_object(), as an example:

$row = mysql_fetch_object($result,'foo');

Note: If your query is going to return more than one row you will have to loop over the $result, which has many examples in the manuals provided.

Note: Please have a look at mysqli overview, which should show you the differences between mysql and mysqli.

edited Dec 30 '17 at 16:41

answered Dec 30 '17 at 16:33

ndrwnaguib

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Thanks,for secure i will learn something new.Now i try to examine your sample and use it. – miro toff Dec 30 '17 at 16:35
You are welcome. Good Luck! – ndrwnaguib Dec 30 '17 at 16:36
Nothing, it show only A not the number picked from DB. – miro toff Dec 30 '17 at 16:40
Are you sure that you are concatenating the `$letter` with `$row`? – ndrwnaguib Dec 30 '17 at 16:42
this code return me the correct value when testing with echo, but when i use it to create database it write Array and not the number.$query = mysqli_query($conn, "SELECT `number` FROM `progress`"); $row = mysqli_fetch_row($query); $try = ($row[0]); echo $try; – miro toff Dec 30 '17 at 18:23
You cannot use `echo` to print arrays, use `print_r($row)` instead. – ndrwnaguib Dec 30 '17 at 18:40

Pick a value from table and put in a variable

1 Answers1