CODEWITHSUNDEEP
phpmysql
Jack
Jack

Reputation: 245

Use a submit button to show data from a database

I am trying to learn php/mysql and working with a project where I want a submit button to show the data I select from a database.

I am for the moment using this statement, and I want it as default.

$result = mysqli_query($con,"SELECT * FROM persons");

But when someone select something else from my "select" I want it to change to something like this:

$result = mysqli_query($con,"SELECT * FROM persons where firstname='$z'");

So when someone select "Charlie" and press "show" I want only data with the firstname Charlie to be shown.

You can follow the project live here: http://www.adamskymotorkylare.se/business/

Thanks in advance Jack

<form action="#" method="post">
<select name="first_name" type="text">
<option value="All">All</option>
<option value="Charlie">Charlie</option>
<option value="Bob">Bob</option>
<option value="Sibylla">Sibylla</option>
</select>
<input type="submit" name="submit" value="Show"/>
</form>

<?php
// Create connection

$z = $_POST['first_name'];

$result = mysqli_query($con,"SELECT * FROM persons");

echo "<table width=100%>
<tr>
<th>Firstname</th>
<th>Lastname</th>
<th>Age</th>
<th>Gender</th>
</tr>";

while($row = mysqli_fetch_array($result))
{

$id = $row['id'];

echo "<tr>";
echo "<td> <a href='view_more.php?id=" . $id ."'>" . $row['firstname'] . "</a> </td>";
echo "<td>" . $row['lastname'] .  "</td>";
echo "<td>" . $row['age'] . "</td>";
echo "<td>" . $row['gender'] . "</td>";
echo "</tr>";
}
echo "</table>";

mysqli_close($con);
?>

Upvotes: 0

Views: 5339

Answers (2)

Simeon
Simeon

Reputation: 408

You could do it like this:

if($_POST['first_name']!="All" && isset($_POST['first_name']))
{
    $firstname=htmlentities($_POST['first_name'], ENT_QUOTES);
    $query="SELECT * FROM persons WHERE firstname='".$firstname."'";
}else
{
    $query="SELECT * FROM persons";
}

echo '
                        <select name="first_name">
                            <option value="All">- - - All - - -</option>
';
$dbPersons=mysqli_query($con,"SELECT DISTINCT firstname FROM persons ORDER BY firstname");
while($DataPersons=mysqli_fetch_array($dbPersons))
{
    $selected='';
    if($DataPersons['firstname']==$firstname)
    {
        $selected=' selected="selected"';
    }
    echo '<option value="'.$DataPersons['firstname'].'"'.$selected.'>'.$DataPersons['firstname'].'</option>';
}
echo '              
                        </select>
';

$result = mysqli_query($con,$query);

echo "
    <table width=100%>
        <tr>
            <th>Firstname</th>
            <th>Lastname</th>
            <th>Age</th>
            <th>Gender</th>
        </tr>
";

while($row = mysqli_fetch_array($result))
{
    $id = $row['id'];

    echo "<tr>";
    echo "<td> <a href='view_more.php?id=" . $id ."'>" . $row['firstname'] . "</a> </td>";
    echo "<td>" . $row['lastname'] .  "</td>";
    echo "<td>" . $row['age'] . "</td>";
    echo "<td>" . $row['gender'] . "</td>";
    echo "</tr>";
}
echo "</table>";

mysqli_close($con);

This way you can select any name in the DB. DISTINCT takes care of duplicates.

Upvotes: 1

Cyclonecode
Cyclonecode

Reputation: 29991

You just have to check your post data:

$query = "SELECT * FROM persons";
if(isset($_POST['first_name'])) {
   $query .= " WHERE firstname='".$_POST['first_name']."'";
}
$result = mysqli_query($con, $query);

Note, that you should try to use a prepared statement for the above to be secure.

Upvotes: 2

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