print out rows from database in a table

Question

Why do I need to use echo to print MYSQL data to my table?

Take a look at my code sample.

what I am trying to do

Select and show all data in the query in a table.

Problem 1

This works: <td><?php echo "<br>". $row["testName"]. "<br>";?></td>

This does not: <td><?php $row["testName"] ?></td>

I feel the second option should work but does not.

(not such a big deal just feels wrong)

Problem 2

I would also like all data to be in one table not for the loop to create a new table every time.

    $sql = "SELECT testName, userID FROM  `results` WHERE  `userID` = '$something' ";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        ?>

        <table class="table table-hover">
            <thead>
            <tr>
                <th>ID</th>
                <th>Name</th>
                <th>some text Score</th>
                <th>Date Taken</th>
            </tr>
            </thead>
            <tbody>
            <tr>
                <th scope="row">1</th>
                <td><?php  $row["testName"] ?></td>
                <td><?php  $row["userID"] ?></td>
                <td><?php echo "<br>". $row["testName"].  "<br>";?></td>
            </tr>
            </tbody>
        </table>
<?php
    }
} else {
    echo "0 results";
}

<?php $row["testName"] ?> does nothing. You're not actually echoing it. If sort-tags are on, or in PHP higher than 5.4.0, you can do <?= $row["testName"] ?>, which echos it. — Qirel
– Qirel, Commented Nov 3, 2016 at 12:16
1. You forgot to echo the variable. 2. Move the first part and the last part of the table out of the while loop (to before and after it). — jeroen
– jeroen, Commented Nov 3, 2016 at 12:16
Change your second option to <td><?= $row["testName"] ?></td> — Mark Baker
– Mark Baker, Commented Nov 3, 2016 at 12:16

Manthan Dave · Accepted Answer · 2016-11-03 12:26:42Z

1

If you want to display value from database then you must need to use echo. because $row["testName"] contains some value but which value it is, that you can see after echoing it.

To display data in single table try below code:

    <?php 
$sql = "SELECT testName, userID FROM  `results` WHERE  `userID` = '$something' ";
      $result = $conn->query($sql); ?>
      <table class="table table-hover">
        <thead>
            <tr>
                <th>ID</th>
                <th>Name</th>
                <th>some text Score</th>
                <th>Date Taken</th>
            </tr>
        </thead>
    <?php

        if ($result->num_rows > 0) {
        // output data of each row
        while($row = $result->fetch_assoc()) {
    ?>
        <tr>
            <th scope="row">1</th>
            <td><?php echo $row["testName"] ?></td>
            <td><?php echo $row["userID"] ?></td>
            <td><?php echo "<br>". $row["testName"].  "<br>";?></td>
        </tr>

    <?php } ?> 
 </table>
   <?php }else 
       {
         echo "0 results";
        }

edited Nov 3, 2016 at 12:26

answered Nov 3, 2016 at 12:19

Manthan Dave

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2 Comments

Benjamin Oats Over a year ago

Plus 1 and check, nice quick fix

Manthan Dave Over a year ago

Thanks Keep helping :)

Waleed Ahmed Haris · Accepted Answer · 2016-11-03 12:18:08Z

1

Run while like this

  <?php
  while($row = $result->fetch_assoc()) { ?>
  <tr>
         <th scope="row">1</th>
         <td><?php  echo $row["testName"] ?></td>
         <td><?php  echo $row["userID"] ?></td>
         <td><?php echo "<br>". $row["testName"].  "<br>";?></td>
  </tr>
  <?php } ?>

answered Nov 3, 2016 at 12:18

Waleed Ahmed Haris

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Comments

Vision Coderz · Accepted Answer · 2016-11-03 12:24:49Z

1

Description

void echo ( string $arg1 [, string $... ] )

Outputs all parameters. No additional newline is appended.

echo is not actually a function (it is a language construct), so you are not required to use parentheses with it. echo (unlike some other language constructs) does not behave like a function, so it cannot always be used in the context of a function. Additionally, if you want to pass more than one parameter to echo, the parameters must not be enclosed within parentheses.

echo also has a shortcut syntax, where you can immediately follow the opening tag with an equals sign. Prior to PHP 5.4.0, this short syntax only works with the short_open_tag configuration setting enabled.

Ref : http://php.net/manual/en/function.echo.php

edited Nov 3, 2016 at 12:24

answered Nov 3, 2016 at 12:17

Vision Coderz

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