Show data based on form submission

Question

0.00/5 (No votes)

See more:

I have a mySQL database and a form page in php. I need the best way to retrieve data from the database based on user input in the form. Now, the form will limit user input to an integer they will be given before hand. The integer will be in the database as an auto increment number that will be called 'rep ID'. The user will input the 'rep ID' to display that rep's information.

I have tried several different approaches, yet have not been able to get a result. This has been ongoing for about a month, with me working on it daily. As of right now, I have the form working the way I want it, the database created, the submission page for collecting the data is perfect... the only problem is I can't manage to pull the data needed based on the user inputting a specific ID.

The code submitted is just the last I have tried. It however, did not work as the several before. Any suggestions would be greatly appreciated, and I just really need a direction to head in. I'm totally lost with this, what I thought was going to be relatively easy, and am growing frustrated. Thank you in advance.

What I have tried:

PHP

<?php 
 $host = 'myhost';
 $user = 'myuser';
 $pass = 'mypassword';
 $db = 'mydatabase';

	$con = mysqli_connect("$host", "$username", "$password") or die("Error connecting to database: ".mysqli_error());
	mysqli_select_db($con, "$db") or die(mysqli_error());

    $query = $_GET['query']; 

	$sql = "SELECT * FROM 'repData' WHERE 'repID' = $query";
	$result = $con->query($sql);

	if ($result->num_rows > 0) {
		echo "";

		while($row = $result->fetch_assoc()) {
			echo "";
		}
		echo "<table><tbody><tr><th>ID</th><th>Name</th></tr><tr><td>".$row["repID"]."</td><td>".$row["repName"]." ".$row["repBio"]."</td></tr></tbody></table>";
	} else {
		echo "<br><h1 align="center">Sorry that ID could not be found in our database.<br>
		Verify the number and try again.</h1>";
	}

	$con->close();
	?>

Posted 3-Apr-17 5:30am

Jawanaut

Updated 5-Apr-17 4:18am

v2

Add a Solution

Comments

W∴ Balboos, GHB 3-Apr-17 12:09pm

Here's a hint for you:

 $array = mysqli_fetch_row ($result);

What is $result?

If you copied this as a sample from somewhere you need to actually use values that match your code - not those of the sample.

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Jochen Arndt · Accepted Answer · 2017-04-05T04:18:00

For version 2 of the question:

PHP

while($row = $result->fetch_assoc()) {
    echo "";
}
echo "<table><tbody><tr><th>ID</th><th>Name</th></tr><tr><td>".$row["repID"]."</td><td>".$row["repName"]." ".$row["repBio"]."</td></tr></tbody></table>";

Print the table header first, then the row contents within the while loop and finally the closing table tags:

PHP

echo "<table><tbody><tr><th>ID</th><th>Name</th></tr>";
while($row = $result->fetch_assoc()) {
    echo "<tr><td>".$row["repID"]."</td><td>".$row["repName"]." ".$row["repBio"]."</td></tr>";
}
echo "</tbody></table>";

Because you are filtering for a single ID, printing the headers outside the while loop here is not really necessary. But you must access $row inside the loop because it is out of scope when using it after the while loop. Even when $row is declared outside the loop if would not work because it is NULL when the loop has terminated.