Adding HTML Button using PHP Mysql

Question

In an application that I am building, I need to display different kinds of buttons for users with different designations (HOD, Director or Staff). Consider the code snippet-

        <button id="request">Leave Request</button>
        <button id="statistics">Leave Statistics</button>
        <?php
        $sql = "SELECT Role from employee where `Emp Code`=$emp_code";
        $result = mysqli_query($link,$sql);
        $row =mysqli_fetch_array($result, MYSQLI_NUM);
        if($row[0] == "HOD"){
            echo "<button id='new'> New Button</button>";
        }

So is the logged in used is a HOD then he/she should see 3 buttons.

Error displaying- Couldn't fetch query. Parameter 1 in mysqli_fetch_array should be mysqli_result, none given.

your query failed and you need to find out why `$result = mysqli_query($link,$sql) or die(mysqli_error($link));` — Funk Forty Niner, Sep 17 '15 at 04:46
`$row =mysqli_fetch_array($result, MYSQLI_NUM);` and you're checking for a string. — Funk Forty Niner, Sep 17 '15 at 04:46
@Fred-ii using die(mysqli_error($link)); , it shows the same error explained above. Is it not the proper way to check for string? How should I check then? — Raghav Motwani, Sep 17 '15 at 04:53
plus, code snippets isn't full code. So, I don't know what else I would be up against. Just do as Bart said, quote your variable; it's a string, not an integer. You may see magic happen ;-) — Funk Forty Niner, Sep 17 '15 at 04:59

Shehary · Answer 1 · 2015-09-17T05:10:08.567

3

As OP mentioned in comments the table structure is emp_name, emp_code, password, emailID, role - varchar(50) and $emp_code is string.

and the query OP is trying;

$sql = "SELECT Role from employee where `Emp Code`=$emp_code";

so query should be

$sql = "SELECT role FROM employee WHERE emp_code='$emp_code'";

edited Sep 17 '15 at 05:10

answered Sep 17 '15 at 05:02

Shehary

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even though I use "SELECT Role from employee where `Emp Code`='aal01' ". aal01 is a value in 'Emp Code' column , it is showing the same error – Raghav Motwani Sep 17 '15 at 05:19
check the queries in answer, you have wrong table col names in query, if as you mentioned in comments that the table col is `emp_code` but in query you are trying with `Emp Code` – Shehary Sep 17 '15 at 05:20
and same goes with col `role` and you are trying `Role` – Shehary Sep 17 '15 at 05:22
1

there is no possible way the col name will be `Emp Code` can't put space between 2 words, so it's totally wrong – Shehary Sep 17 '15 at 05:23
@Shehary Not quite ^ ^ Although not considered good practice, spaces are allowed but ticks are required to be used around table/column names. However, it looks like the OP's column name is different than what they originally used. – Funk Forty Niner Sep 17 '15 at 12:56
@Fred-ii- you are right, actually i never saw and never tried to create a col having space between 2 words, always used `_` so i didn't know that it's possible and it can be done but just created a col with space between 2 words and yes I was Totally Wrong about my comment :) – Shehary Sep 17 '15 at 13:00
@Shehary Same thing for a hyphen `table or col-name`, allowed, but not good practice and also requires ticks around it. MySQL sees that as wanting to do math. I.e.: "col **minus** name". – Funk Forty Niner Sep 17 '15 at 13:02
1

@Fred-ii- yeah agree with you `-` hyphen is not practice also. space between 2 words in col name was something new to me, glad learned something. – Shehary Sep 17 '15 at 13:29

score 1 · Answer 2 · answered Sep 17 '15 at 05:00

I'm on my phone but since emp_code is a string please try the following. Notice the ' ' around $emp_code

        <button id="request">Leave Request</button>
    <button id="statistics">Leave Statistics</button>
    <?php
    $sql = "SELECT Role from employee where `Emp Code`='$emp_code'";
    $result = mysqli_query($link,$sql);
    $row =mysqli_fetch_array($result, MYSQLI_NUM);
    if($row[0] == "HOD"){
        echo "<button id='new'> New Button</button>";
    }

dhi_m · Answer 3 · 2015-09-17T05:07:44.970

0

Please try this .

        <button id="request">Leave Request</button>
        <button id="statistics">Leave Statistics</button>
        <?php
        $sql = "SELECT `role` from employee where `emp_code`='".$emp_code."'";
        $result = mysqli_query($sql);
        $row =mysqli_fetch_array($result, MYSQLI_NUM);
        if($row[0] == "HOD"){
            echo "<button id='new'> New Button</button>";
        }

edited Sep 17 '15 at 05:07

answered Sep 17 '15 at 04:48

dhi_m

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Didn't work even now! – Raghav Motwani Sep 17 '15 at 04:52
Please post `employee` table structure. – dhi_m Sep 17 '15 at 04:54
emp_name, emp_code, password, emailID, role - varchar(50) – Raghav Motwani Sep 17 '15 at 05:00
Please check updated answer.. use emp_code instead of Emp Code.Modified $sql query as per your table structure $sql = "SELECT `role` from employee where `emp_code`=$emp_code"; – dhi_m Sep 17 '15 at 05:08

Adding HTML Button using PHP Mysql

3 Answers3