How do I display my mysql table column headers in my php/html output?
Question
2 Questions...
Scenario:
I would like to query my database table via a form and then display the results that occur (if there are results) and my current situation is that it does work but it clears the form completely and leaves my to an empty page with just the results if that makes sense. I would like to see my results on the same page that I'm entering the form data in. I would like to see my data in a table based format, kind of like this...
|Name | Age|
|-------|----|
|Anthony| 20 |
I have a separate design to make it look pretty later on. This is my setup so far....
displayform.html:
<html>
<form method="post" name="display" action="display.php" />
Enter the name you like to display the data from MySQL:<br>
<input type="text" name="name" />
<input type="submit" name="Submit" value="display" />
</form>
</html>
and display.php
<?php
mysql_connect("localhost", "toor", "FakePassword") or die("Connection Failed");
mysql_select_db("FakeDatabase")or die("Connection Failed");
$name = $_POST['name'];
$query = "select name, age from test WHERE name = '$name'";
$result = mysql_query($query);
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "<tr>";
echo "<td>".$line['name']."</td>";
echo "<td>".$line['age']."</td>";
echo "<br>\n";
echo "</tr>";
}
?>
2 answers
solution1
2 ACCPTED 2015-06-22 02:51:16
Note:
- You can just make a single file out of it to achieve your wanted output
- Use mysql_real_escape_string() to sanitize the passed-on value to prevent SQL injections
- You should use
mysqli_*instead of the deprecatedmysql_*API
Form them in a single file like this ( display.php ):
<html>
<form method="post" name="display" action="display.php" />
Enter the name you like to display the data from MySQL:<br>
<input type="text" name="name" />
<input type="submit" name="Submit" value="display" />
</form>
<?php
mysql_connect("localhost", "toor", "FakePassword") or die("Connection Failed");
mysql_select_db("FakeDatabase")or die("Connection Failed");
if(!empty($_POST["name"])){ /* WE ADD THIS PART SO WHEN NO PASSED DATA IS FOUND, IT WILL NOT GENERATE "UNIDENTIFIED VARIABLE" ERROR */
$name = mysql_real_escape_string($_POST['name']); /* SANITIZE THE VALUE OF THIS VARIABLE */
$query = "select name, age from test WHERE name = '$name'";
$result = mysql_query($query);
echo "<table>";
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "<tr>";
echo "<td>".$line['name']."</td>";
echo "<td>".$line['age']."</td>";
echo "<br>\n";
echo "</tr>";
} /* END OF WHILE LOOP */
echo "</table>";
} /* END OF NOT EMPTY NAME */
?>
</html>
If you're interested with mysqli_* , I recommend that you use prepared statement . Just replace your php part to this:
<?php
/* ESTABLISH YOUR CONNECTION FIRST */
$con = new mysqli("localhost", "toor", "FakePassword", "FakeDatabase");
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
if(!empty($_POST["name"])){
if($stmt = $con->prepare("SELECT name, age FROM test WHERE name = ?")){
$stmt->bind_param("s",$_POST["name"]); /* BIND THE PASSED-ON VALUE TO THE QUERY */
$stmt->execute(); /* EXECUTE THE QUERY */
$stmt->bind_result($name,$age); /* BIND THE RESULT TO THESE VARIABLES */
echo "<table>";
while($stmt->fetch()){ /* FETCH ALL RESULTS */
echo "<tr>";
echo "<td>".$name."</td>";
echo "<td>".$age."</td>";
echo "<br>\n";
echo "</tr>";
} /* END OF WHILE LOOP */
echo "</table>";
$stmt->close();
} /* END OF PREPARED STATEMENT */
} /* END OF NOT EMPTY NAME */
?>
solution2
0 2015-06-22 02:57:55
You must use this query if you want the columns names: SHOW COLUMNS FROM test
This is a working example.
Also, be careful with SQL injection . And if you didn't receive it, my web server thrown a warning into my screen:
The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead
<?php
mysql_connect("localhost", "toor", "FakePassword") or die("Connection Failed");
mysql_select_db("FakeDatabase")or die("Connection Failed");
$query = "SHOW COLUMNS FROM test";
$result = mysql_query($query);
echo "<tr>";
while ($header = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "<td>". $header['Field']."</td>";
}
echo "<br>\n";
echo "</tr>";
$name = $_POST['name'];
$query = "select name, age from test WHERE name = '$name'";
$result = mysql_query($query);
while ($line = mysql_fetch_array($result, MYSQL_ASSOC)) {
echo "<tr>";
echo "<td></td>";
echo "<td>".$line['name']."</td>";
echo "<td>".$line['age']."</td>";
echo "<br>\n";
echo "</tr>";
}
?>
EDIT
I answered the main question as described in the question title. Next time, you should post one question... per question. Good luck.
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