How to show mysql multi row / mysql_fetch_array results in a table?

Question

I am trying to show mysql_fetch_array() results in a table.

I want to show guests name,their country and their agreed time who are traveling on a same date.

Following code works fine. The code fetches the row values and prints it.

 $select_guests = mysql_query('SELECT name FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error()); // query for getting guests for the same date
while($row = mysql_fetch_array($select_guests, MYSQL_ASSOC)) { //visitor / guest loop starts here
    echo $row['name'].'<br/>';
}
$select_country = mysql_query('SELECT country FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error()); // query for getting guests for the same date
while($row = mysql_fetch_array($select_country, MYSQL_ASSOC)) { //country of visitor / guest loop starts here
    echo $row['country'].'<br/>';
}
$select_agreed_time = mysql_query('SELECT agreed_time FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error()); // query for getting guests for the same date
while($row = mysql_fetch_array($select_agreed_time, MYSQL_ASSOC)) { //visitor / guest agreed time loop starts here
    echo $row['agreed_time'].'<br/>';
}

if there are 5 guests for a same date I am getting all of their names one below another when I execute the above code. Same I am getting there countries and agreed time too.

Now I want to show those results in a HTML table. I tried several line of code but nothing works.

My HTML table should be as following:

<table class="table-fill">
    <thead>
        <tr>
            <th class="text-left">Name</th>
            <th class="text-left">From</th>
            <th class="text-left">Agreed Time</th>
        </tr>
    </thead>
    <tbody class="table-hover">
        <tr>
            <td class="text-left">Name 1</td>
            <td class="text-left">Country 1</td>
            <td class="text-left">Ag Time 1</td>
        </tr>
        <tr>
            <td class="text-left">Name 2</td>
            <td class="text-left">Country 2</td>
            <td class="text-left">Ag Time 2</td>
        </tr>
        <tr>
            <td class="text-left">Name 3</td>
            <td class="text-left">Country 3</td>
            <td class="text-left">Ag Time 3</td>
        </tr>
        <tr>
            <td class="text-left">Name 4</td>
            <td class="text-left">Country 4</td>
            <td class="text-left">Ag Time 4</td>
        </tr>
        <tr>
            <td class="text-left">Name 5</td>
            <td class="text-left">Country 5</td>
            <td class="text-left">Ag Time 5</td>
        </tr>
    </tbody>
</table>

How can create that table td s according to my mysql_fetch_array() ? The above table structure is for 5 guests found or resulted by mysql_fetch_array()

`mysql` extension is deprecated, consider using `mysqli` or `PDO` — Muhammet, Jun 18 '15 at 11:27
@Muhammet thanks.. noted.. **DOWN VOTERS** : I am asking here when I do not understand / dont know somthing about.. so when someone really want to **DOWN VOTE** please comment the reason at least.. I am NOT genius enough to understand why I got down vote.. — Foolish Coder, Jun 18 '15 at 11:40
If you can, you should [stop using `mysql_*` functions](http://stackoverflow.com/questions/12859942/why-shouldnt-i-use-mysql-functions-in-php). They are no longer maintained and are [officially deprecated](https://wiki.php.net/rfc/mysql_deprecation). Learn about [prepared](http://en.wikipedia.org/wiki/Prepared_statement) [statements](http://php.net/manual/en/pdo.prepared-statements.php) instead, and consider using PDO, [it's really not hard](http://jayblanchard.net/demystifying_php_pdo.html). — Jay Blanchard, Jun 18 '15 at 12:10

habib ul haq · Accepted Answer · 2015-06-18T11:42:28.483

First of all I think you dont need 3 different queries for your solution..

    <table class="table-fill">
            <thead>
                <tr>
                    <th class="text-left">Name</th>
                    <th class="text-left">From</th>
                    <th class="text-left">Agreed Time</th>
                </tr>
            </thead>
        <?php 
        $result = mysql_query('SELECT name,country,agreed_time FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error()); 
        while($row = mysql_fetch_array($result, MYSQL_ASSOC)) 
        { 
        ?>
          <tr>
               <td>
                   <?php    echo $row['name']; ?>
               </td>
               <td>
                   <?php     echo $row['country'];?>
               </td>
                <td>
                   <?php     echo $row['agreed_time']; ?>
               </td>    
    </tr>
    <?php
    }
?>
</table>

score 1 · Answer 2 · answered Jun 18 '15 at 11:44

Firstly, you should use mysqli.

Secondly, shouldn't be sending so many queries to the database for information you can get in one query;

$select_guests = $mysqli->query('SELECT * FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error());

Next, you want to fetch the number of rows.

$rows = $mysqli

You should also look into the PHP for function;

for ($i = 1; $i < $rows; $i++) {
    $thisRow = $select_guests->fetch_row()
    echo
'        <tr>
        <td class="text-left">'.$select_guests['name'].'</td>
        <td class="text-left">'.$select_guests['country'].'</td>
        <td class="text-left">'.$select_guests['time'].'</td>
    </tr>
    '; //This last line is to insert a line break and indent (for tidy HTML)
}

Give this a go, hopefully I've helped you. I haven't completely solved it for you though, in order to change over to mysqli, you will need to make a few small changes which you can find in the mysqli link I sent you. The benefits are worth it.

Thanks for this.. I'll try it another time.. I am already using Habib Ul haq's suggestion.. I upvoted you.. your answer is tidy and neat.. — Foolish Coder, Jun 19 '15 at 05:02
Thankyou. I appreciate the up vote and your response. Best of luck. — Michael Thompson, Jun 19 '15 at 05:06

Kiran Dash · Answer 3 · 2015-06-18T11:44:35.507

0

$select_all = mysql_query('SELECT * FROM van_sharing WHERE date = "'.$serch_text.'"') or die(mysql_error()); // query for getting all details for the same date
while($row = mysql_fetch_array($select_all , MYSQL_ASSOC)) {//loop starts here
    <tr>
           <td>
               <?php    echo $row['name']; ?>
           </td>
           <td>
               <?php     echo $row['country'];?>
           </td>
            <td>
               <?php     echo $row['agreed_time']; ?>
           </td>    
  </tr>

}

edited Jun 18 '15 at 11:44

answered Jun 18 '15 at 11:33

Kiran Dash

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this doesnt answer the question. the OP wants to populate the rows of a table. – DevDonkey Jun 18 '15 at 11:36
I think that gives the idea. But i still made the edit – Kiran Dash Jun 18 '15 at 11:44
Why should the OP do this? Please add an explanation of what you did and why you did it that way, not only for the OP but for future visitors to SO. – Jay Blanchard Jun 18 '15 at 12:13

How to show mysql multi row / mysql_fetch_array results in a table?

3 Answers3