Select statement in php function

Question

I am attempting to create a function that will take a query in mysql and output it with mysql_fetch_assoc.

The issue is when I return the output inside the loop it only outputs the first field. Likewise, if I return the output outside of the loop I only get the last value.

Is there anyway to get around this?

Here is the main page:

<?php
$dbhost = "localhost";
$dbuser = "widget_cms";
$dbpass = "cjclone123";
$dbdata = "widget_corp";
$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbdata);

if(mysqli_connect_errno()) {
    die("database connection failed: " .
        mysqli_connect_error() .
            " (" . mysqli_connect_errno() . ")"
    );
} else {
    echo "success!";
} 

Insert_Line_Into_subjects();

    $table = "subjects";

    $result = Output_Table($table);

    var_dump($result);

    echo (Display_Table($result));

    mysqli_free_result($result); 


?>  
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"
   "http://www.w3.org/TR/html4/loose.dtd">

<html lang="en">
 <head>
  <title>phpMysql</title>
 </head>
 <body>

    <ul>
     <form action="phpMysql.php" method="post">
     Menu Name: <input type="text" name="menuName" value=""><br>
     Position: <input type="text" name="pos" value=""><br>
     Visibility: <input type="text" name="vis" value=""><br>
     <input type="submit" name="submit" value="Create account"><br>
     </form>
    </ul> 

 </body>
</html>

<?php mysqli_close($connection); ?>

Here are all the functions that I used:

function Insert_Line_Into_subjects(){

if (isset($_POST["submit"])) {
    global $connection; 

    $menu_name = ($_POST["menuName"]);
    $position = ($_POST["pos"]);
    $visibility = ($_POST["vis"]);

    $query1 = "INSERT INTO subjects ";
    $query1 .= "(menu_name, position, visible) ";
    $query1 .= "VALUES ('$menu_name', '$position', '$visibility')";
    $result = mysqli_query($connection, $query1);
    Check_SQL_Execution($result);
    return $result;
    }
}

    function Check_SQL_Execution($result) {
        global $connection;

    if (!$result) {
        die("<br>Database query failed: " . mysqli_error($connection));
    }
    }

    function Output_Table ($table) {
    global $connection; 

    $query1 = "SELECT * ";
    $query1 .= "FROM {$table}" ;
    $result1 = mysqli_query($connection, $query1);
    Check_SQL_Execution($result1);
    return $result1; 

}

    function Display_Table($result) {

        //if (isset($_POST["submit"])){
        $output = '';

        while ($row = mysqli_fetch_assoc($result)) {
        $output = $row["menu_name"];
        $output .= " ("; 
        $output .= $row["id"];
        $output .= ")";
    }
    return $output;
 }

What did I do wrong? Thanks!

I expected the entire table to output. Each menu name and id. — zach
– zach, Commented Jun 1, 2016 at 13:45

Alex · Accepted Answer · 2016-06-01 13:36:41Z

4

function Display_Table($result) {
    $output = '';
    while ($row = mysqli_fetch_assoc($result)) {
        $output .= $row["menu_name"];
        $output .= " ("; 
        $output .= $row["id"];
        $output .= ")";
    }
    return $output;
}

answered Jun 1, 2016 at 13:36

Alex

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4 Comments

zach Over a year ago

I tried this and it only outputted the last value in the table.

Alex Over a year ago

show me the proof. show me full fragment of your code (where and how you use this function), query and sample of data you execute your query against

Progrock Over a year ago

@zach, the subtle difference here with Alex's snippet is that $output is appended to on each loop, whereas you were clobbering it each time.

zach Over a year ago

That was it! Thank you.

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