Not giving the wanted output

Question

So I'm having this issue with showing the output of a query. I will show my code and try to explain the best I can.

<?php
    include_once('ligacao.php');
    if (mysqli_connect_errno())
    {
    echo "Falha ao conectar à Base de Dados: ".mysqli_connect_error();
    }

        $pesq = $_POST['id'];
        echo $pesq; 

        $sqlveri = "SELECT * FROM distrito WHERE iddistrito LIKE '12'";
        $result = mysql_query ($ligar,$sqlveri);
        echo $resul;
        mysqli_close ($ligar);
        ?>

The only thing that appears when I submit is the number that is kept on the variable $pesq. Help me out here guys I have been grinding for an answer for 2 straight days with no luck at all

Thank you for your help.

so you changed `mysql_fetch_array` to `mysqli_fetch_array`. How do we know you're not using `mysql_` to connect with or PDO? Plus, you're not checking for errors. Also, no idea what variable you are using to connect with. — Funk Forty Niner, Dec 17 '15 at 13:02
I reopened your question and I have posted my answer for you below. Edit: you deleted your comment that I replied to here. — Funk Forty Niner, Dec 17 '15 at 13:18
you just ***completely changed*** your question from your original post and http://stackoverflow.com/revisions/34335181/2 - I won't be offering any further help on this, sorry. You need to read more tutorials and manuals. Had I known this ahead of time, I'd of never submitted an answer. — Funk Forty Niner, Dec 17 '15 at 13:35
and this is wrong `$result = mysql_query ($ligar,$sqlveri); echo $resul;` I have done what I could here. Good luck with your project. — Funk Forty Niner, Dec 17 '15 at 13:38

score 5 · Answer 1 · edited May 23 '17 at 11:44

Note to future visitors.

The original closure of their question was based on their original post where they were using mysql_fetch_array without the i, and changed it before it could be recorded in revisions.

Then completely changed their code https://stackoverflow.com/revisions/34335181/3 after I posted my answer.

Original answer based on the above

So you changed mysql_fetch_array to mysqli_fetch_array after I closed the question about your mixing MySQL APIs.

It is unclear as to which MySQL API you're using to connect with. mysql_, mysqli_, or PDO (those different APIs do not intermix).

Can I mix MySQL APIs in PHP?

Plus, you're not checking for errors.

Also, no idea what variable you are using to connect with.

http://php.net/manual/en/function.mysqli-connect.php

What's going on here is that you're not connecting to your query and I do not know what variable you are using.

So, base yourself on the following and change the variable to the one you're using in your connection file:

$result = mysqli_query($connection, "SELECT ...

mysqli_query() requires a connection:

http://php.net/manual/en/mysqli.query.php

Add error reporting to the top of your file(s) right after your opening PHP tag for example <?php error_reporting(E_ALL); ini_set('display_errors', 1); then the rest of your code, to see if it yields anything, as well as or die(mysqli_error($connection)) to mysqli_query().

Your HTML form is also unknown, and if it is using a POST method and if the input does have the name attribute for it.

I.e.:

<form method="post" action="your_handler.php">
  <input type="text" name="id">
</form>

Forms default to a GET if a POST method isn't specifically set. Therefore, that would trigger/cause an undefined index notice.

Sidenote: Make sure there are no whitespaces in your input/query. Use trim() and var_dump() as an additional tool.

Plus, if you're looking for an exact match, don't use LIKE, but a WHERE iddistrito = '$pesq'");

Read up on LIKE: http://dev.mysql.com/doc/refman/5.7/en/pattern-matching.html

There are wildcards you can use, if pattern matching is required.

Your present code is open to SQL injection. Use mysqli_* with prepared statements, or PDO with prepared statements.

Great explanation Fred you've touched every point that is wrong with his code. I have two little question: If I enable `error_reporting(E_ALL);` how will it display without me specifying it at a specific moment in the code like `mysqli_error($connection)` for example? second do I still have to use `mysqli_error($connection)` if I enable `error_reporting(E_ALL);`? — BRoebie, Dec 17 '15 at 13:27
@BRoebie Thanks. only using `error_reporting(E_ALL);` will not display errors, but log them, therefore using `ini_set('display_errors', 1);` is best to display them if people don't have access to logs. — Funk Forty Niner, Dec 17 '15 at 13:29
thank you for your fast reply so if I am understanding it correctly if I use this ` error_reporting(E_ALL); ` combined with this `ini_set('display_errors', 1);` I don't have to use any other error checking in for example the connection string or the query with `mysqli_error()`? — BRoebie, Dec 17 '15 at 13:33
@BRoebie you're welcome. It's best to check for errors for both PHP and MySQL while testing. Then, to remove them for production once everything works. — Funk Forty Niner, Dec 17 '15 at 13:35
Okay I understood all of the points you refered. I have the connections there, I changed it for the sake of "try everything". The mysqli's go for the same. You are very correct about the LIKE in the mysql statement. I have already changed it. — picaluga, Dec 17 '15 at 13:38
@Fred-ii- Thanks for the detailed explanation sorry if I sounded like a bit of a stupid idiot. But I wanted to achieve that if somebody came by who just started learning PHP he/she would understand why you would need that to do :D — BRoebie, Dec 17 '15 at 14:38
@BRoebie You're welcome. I can only help with what I see before me. You're not a *"stupid idiot"*, you're learning and that doesn't make you "stupid". I tried helping the best I could with what I had to work with. If there's something that's unclear/unknown somewhere, then I can't help there since I don't know what I'm up against ;-) I wish you all the very best, *cheers* — Funk Forty Niner, Dec 17 '15 at 14:41

Not giving the wanted output

1 Answers1