asked on

Insert values are dynamic

I have this code:

<?php
require "connection.php";
error_reporting(E_ALL);
ini_set('display_errors',1);
$username = "jasmine";
?>
<!DOCTYPE html>
<html lang="en">
<head>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width, initial-scale=1">
  <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
  <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
  <link href="//netdna.bootstrapcdn.com/bootstrap/3.0.3/css/bootstrap.min.css" rel="stylesheet">
<link href = "http://fonts.googleapis.com/css?family=Roboto:400">
<link href="//maxcdn.bootstrapcdn.com/font-awesome/4.2.0/css/font-awesome.min.css" rel="stylesheet">
<style>
.modal-dialog{
	width:100%;
	overflow: scroll;
	
}

</style>


</head>
<body>


  <h2>Modal Example</h2>
  <!-- Trigger the modal with a button -->
  <button type="button" class="btn btn-info btn-lg" data-toggle="modal" data-target="#myModal">Open Modal</button>

  <!-- Modal -->
  <div class="modal fade" id="myModal" role="dialog">
    <div class="modal-dialog">
    
      <!-- Modal content-->
      <div class="modal-content">
        <div class="modal-header">
          <button type="button" class="close" data-dismiss="modal">&times;</button>
          <h4 class="modal-title">ADD</h4>
        </div>
        <div class="modal-body">
          <p>


<form action="" method="post">
<table>
<thead>
   <?php 
    $query = mysqli_query($conn,"SHOW columns FROM `rsvp`");
    // Loop over all result rows
    while($row = mysqli_fetch_array($query))
    {
        echo '<th>';
        echo $row["Field"];
        echo '</th>';
    }
 ?>
 </thead>
 <tbody>
 <?php 
    $query4="SELECT * from `rsvp` WHERE `Gathered By` = '$username'";

    $result4=  mysqli_query($conn, $query4);
    while($row4 = mysqli_fetch_assoc($result4)){
        echo '<tr>';
        foreach($row4 as $values3){
        echo '<td>';
        echo '<input type="text" name="' .$row['Field'] .'">';
        echo '</td>';
        }
        ?>
        <td>
        <input type="submit" name="add" value="Add">
        </td>
        <?php
        echo '</tr>';}
 ?>
 </tbody>
</table>
</form>
          
<?php
if(isset($_POST['add'])){
      
$result2= mysqli_query($conn,"SELECT * FROM users WHERE username = '$username'");
$row2 = mysqli_fetch_row($result2);

$row['Field'] = $columns;
mysqli_query($conn,'INSERT INTO table (text, category) VALUES '.implode(',', $columns));




//// WHAT DO I DO HERE?

}
?>  
          
          </p>
        </div>
      </div>
      
    </div>
  </div>
  


</body>
</html>

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As you can see it gets me the column names of my table along with the text for the username ="jasmine", I want to be able to allow users to add another row to the columns and for the "gathered by" area to always be static which is jamine or whatever the name is of the person logged in. I think I got most of the steps down, Im just stuck on the insert query since I won't know any static names, is there a way to do it even without me knowing the column name?
Right now I have these errors:
Notice: Undefined variable: columns

Notice: Undefined variable: columns

Warning: implode(): Invalid arguments passed

Ray Paseur

Lines 69-74... Try something more like this. You want column names, right?

    while($row4 = mysqli_fetch_assoc($result4)){
        echo '<tr>';
        foreach($row4 as $key => $values3){
        echo '<td>';
        echo '<input type="text" name="' .$row4[$key] .'" />';
        echo '</td>';

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Jazzy 1012

ASKER

Yeah But im still getting the same errors I think I have something wrong in the query part.

<?php
if(isset($_POST['add'])){
      
$result2= mysqli_query($conn,"SELECT * FROM users WHERE username = '$username'");
$row2 = mysqli_fetch_row($result2);

$row["Field"] = $columns;
mysqli_query($conn,'INSERT INTO table (text, category) VALUES '.implode(',', $columns));




//// WHAT DO I DO HERE?

}
?>

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Ray Paseur

First, we need to fix this. In the snippet, $columns is an undefined variable, and $row["Field"] is an unused variable. You do not want to have extraneous code and variables in your scripts.

$row["Field"] = $columns;

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When your script runs a query, you can test for success or failure, and print out the error messages. Follow the code examples in this article.
https://www.experts-exchange.com/articles/11177/PHP-MySQL-Deprecated-as-of-PHP-5-5-0.html
Look in the article for Run the CREATE TABLE Query and Handle Errors/Exceptions. You will need to make some adjustments because you're using procedural code, but you can still use the error testing and visualization techniques.

Julian Hansen

There are many things I don't understand about this code
1. Your loop to output the headers has this

while($row = mysqli_fetch_array($query))

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Then later in your code to output fields you reference $row["Field"] but this will now only refer to the last $row from the header loop which will have exhausted the array and left $row empty.

2. Consider this line.

        echo '<input type="text" name="' .$row['Field'] .'">';

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In the above snippet you are in a loop - each time it goes round name will be set to $row["Field"] which as a loop invariant - it is not altered in this loop so will be the same value. What you probably wanted was

    while($row4 = $result4->fetch_assoc()){
        echo '<tr>';
        foreach($row4 as $field=>$values3){
        echo '<td>';
        echo '<input type="text" name="' .$field .'">';
        echo '</td>';
        }

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But event this does not make sense because if your query returns more than one row you will have multiple <input> controls with the same name

Can you explain a bit more about what you are expecting this page to do.

Jazzy 1012

ASKER

I want this page to display all the columns of the table called "rsvp" and with one input box under each column (therefore just added on row), and after I write the information and I click add, it inserts the information in my database depending on the column it is on, it is similar to my editing field. But how can I do it with insert?

Julian Hansen

Will there only be one row - or is it possible to have multiple records returned?

Jazzy 1012

ASKER

Here I can make it more clear:
in page 1 have this code:

<?php
require "connection.php";
error_reporting(E_ALL);
ini_set('display_errors',1);
$username = "jasmine";
?>
<!DOCTYPE html>
<html lang="en">
<head>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width, initial-scale=1">
  <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
  <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.1.1/jquery.min.js"></script>
  <script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
  <link href="//netdna.bootstrapcdn.com/bootstrap/3.0.3/css/bootstrap.min.css" rel="stylesheet">
<link href = "http://fonts.googleapis.com/css?family=Roboto:400">
<link href="//maxcdn.bootstrapcdn.com/font-awesome/4.2.0/css/font-awesome.min.css" rel="stylesheet">
<style>
.modal-dialog{
	width:100%;
	overflow: scroll;
	
}

</style>


</head>
<body>


  <h2>Modal Example</h2>
  <!-- Trigger the modal with a button -->
  <button type="button" class="btn btn-info btn-lg" data-toggle="modal" data-target="#myModal">Open Modal</button>

  <!-- Modal -->
  <div class="modal fade" id="myModal" role="dialog">
    <div class="modal-dialog">
    
      <!-- Modal content-->
      <div class="modal-content">
        <div class="modal-header">
          <button type="button" class="close" data-dismiss="modal">&times;</button>
          <h4 class="modal-title">ADD</h4>
        </div>
        <div class="modal-body">
          <p>


<form action="try2.php" method="post">
<table>
<thead>
   <?php 
    $query = mysqli_query($conn,"SHOW columns FROM `rsvp`");
    // Loop over all result rows
    while($row = mysqli_fetch_array($query))
    {
     if($row["Field"] == 'id')
     {
			continue;
	 }   
		echo '<th>';
        echo $row["Field"];
        echo '</th>';
    }
 ?>
 </thead>
 <tbody>
 <?php 
    $query4="SELECT * from `rsvp` WHERE `Gathered By` = '$username'";

    $result4=  mysqli_query($conn, $query4);
  while($row4 = $result4->fetch_assoc()){
        echo '<tr>';
        foreach($row4 as $field=>$values3)
        {
			if($field == 'id')
			{
				continue;
			}
        	echo '<td>';
        	echo '<input type="text" name="' .$field .'">';
        	echo '</td>';
        }
        ?>
        <td>
        <input type="submit" name="add" value="Add">
        </td>
        <?php
        echo '</tr>';}
 ?>
 </tbody>
</table>
</form>

        
        </div>
      </div>
      
    </div>
  </div>
  


</body>
</html>

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In this code, I get the column names along with the input text under each so I can insert the information, after I click add, it goes to this page:

<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
session_start();
$username= $_SESSION['username'];
if($_SESSION['username'] == "")
{
	header("Location: http://markitlive.com/users/");
}

require "connection.php";

if(isset($_POST['add'])){
      
$result2= mysqli_query($conn,"SELECT * FROM users WHERE username = '$username'");
$row2 = mysqli_fetch_row($result2);

$field= $columns;
mysqli_query($conn,'INSERT INTO table (text, category) VALUES '.implode(',', $columns));




//// WHAT DO I DO HERE?

}
?>

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Where I insert the information to the database

Jazzy 1012

ASKER

One row, do you want to see an image of the page?

Jazzy 1012

ASKER

Also can I have the input box for the name already prefilled with the username and the date prefilled with todays date?

Jazzy 1012

ASKER

I did the date and username, but how can I do the insert?

ASKER CERTIFIED SOLUTION