I'm struggling with a strange error.
I create an HTML table from a mysql query using PHP like so:
echo "<div id=\"table_div\">";
echo "<table id=\"ad_table\">";
echo "<tr id=\"table_header\"><td>Media Id</td><td> Company</td><td>Title/Claim</td><td>Start - End</td><td>Website/Email</td><td></td></\td></tr>";
while($row = mysql_fetch_assoc($result)) {
echo "<tr id= " . $row['Id']. "><td><b>" .$row['Titel'] . "</b> <br>" . $row['Offer'] . "</td><td><button class=\"del_btn\" rel=" . $id . ">Delete</button> </td></\td></tr>";
}
echo "</table>";
echo "</div>";
But the table is not visible. I can inspect the HTML and table and all expected rows are present but the table class has the attribute display: none. The style is not coming from my stylesheet (I tried it without providing any). I don't know why the table is hidden. Happens in both Chrome and Firefox.
Any help is greatly appreciated.
This is caused by AdBlock browser plugin
Related
I'm trying to create a button for each row in my database that, when pressed, will delete this particular row. I should also mention that the data from the database is displayed correctly and the table I'm using is also completely fine.The buttons appear at the side of each row, when the button is clicked, the row dissapears but the data is not deleted from the database, when the page is reloaded the rows that were previously "deleted", reappear. After pressing the button i also get this "Fatal error: Uncaught Error: Call to undefined function mysql_query() in C:\xampp\htdocs\INDUSTRIALPROJECT\records.php:56 Stack trace: #0 {main} thrown in C:\xampp\htdocs\INDUSTRIALPROJECT\records.php on line 56".
line 56 is : $del = mysql_query("DELETE FROM records WHERE id=" . $row['id']);. The same query works fine when placed directly into phpMyAdmin.
<?php
// Check connection
include_once 'config.php';
if ($link->connect_error) {
die("Connection failed: " . $link->connect_error);
}
$sql = "SELECT * FROM records";
$result = $link->query($sql);
function post($key){ return(isset($_POST[$key]) ? htmlspecialchars($_POST[$key]) : ""); }
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
if(post('rowButton'.$row['id']) =="Delete"){
$del = mysql_query("DELETE FROM records WHERE id=" . $row['id']);
$deleted = '<p>Entry ' . $row['id'] . ' was succesfully deleted</p>';
}
else {
echo '<form action ="' . $_SERVER['PHP_SELF'] . '" method="post">';
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row["visitingdate"]. "</td>";
echo "<td>" . $row["department"] . "</td>";
echo "<td>" . $row["visitingreason"]. "</td>";
echo "<td>" . $row["importance"]. "</td>";
echo "<td>" . $row["visitorname"]. "</td>";
echo "<td>" . $row["company"]. "</td>";
echo "<td>". $row["internalrecipientname"]. "</td>";
echo "<td>". $row["visitinglocation"]. "</td>";
echo "<td>". $row["ETA"]. "</td>";
echo "<td>". $row["ETD"]. "</td>";
echo "<td>". $row["HRverification"]. "</td>";
echo "<td>". $row["visitcompleted"]. "</td>";
echo '<td><input type="submit" name="rowButton'. $row['id'] .'" value="Delete"/> </td>';
echo "</tr>";
echo "</form>";
}
}
echo "</table>";
echo $deleted;
}
else { echo "0 results"; }
$link->close();
?>
First, this appears to be rather vulnerable to SQL injection attacks. StackOverflow is quite font of pointing this out up front, because it's really a solved problem that you should account for in the early stages of development. You're taking untrusted data (that was submitted by the user, without sanitizing it) and putting it directly in an SQL query. Bad things can happen when that occurs. Now with that aside, on to your actual question.
"Nothing happens" means the page doesn't change at all, right? So the browser doesn't know what to do when the button is clicked.
I think you haven't put any <form...> declaration here, which would be required for <input type="submit"> to do anything useful. You could use JavaScript with the stand alone submit button, but I don't see that in your code, either. You'll need something to tell the browser what to do when the submit button is pressed.
I haven't really tested the rest of your code, but based on what you've got already you might add the following:
else {
+ echo '<form action ="' . $_SERVER['PHP_SELF'] . '" method="post">';
echo "<tr>";
and
echo "</tr>";
+ echo "</form>";
}
(don't add the plus sign, that's just to show which line is added). I should add that I don't usually use submit buttons like this, so there's a chance I missed some additional details about how you're calling this, but putting the form in a <form> tag is at least a good start.
Edit
The mysql_query() function was removed in PHP 7; if you're using an older PHP you need to add support for the MySQL functions or if you're on PHP 7, you should use the MySQLi or PDO_MySQL functions instead. The warning box on the PHP manual page for mysql_query has some links for alternatives, how to select an alternative, and other supporting documentation that may help you. This StackOverflow answer may help, as well.
I was wondering how I could add a scroll to my table that is written in the php file. I do not want to write it in a style.css file, I want it directly in the php file. below is my code, but I am not able to make it to work. The table gets content from mySql database, which works. but the problem is that I get to much of content so it fills out the whole page. That is why I want to make it scrollable :
if(mysqli_num_rows($result) > 0){
echo '<table border="1">';
echo "<tr>";
echo "<th>Name</th>";
echo "</tr>";
while($row = mysqli_fetch_array($result)){
echo "<tr>";
echo "<td>" . $row['name'] . "</td>";
echo "</tr>";
}
echo "</table>";
If im getting your question right you need to set a width and height then set the overflow-y. This will give you a table set to a certain size with a scroll bar. Note you need to set your own width and height.
echo "<table style=\"width:500px; height:500px; overflow-y:auto\">";
So currently, I'm making a music database online. The part I'm trying to implement is a playlist function so that given a list of your playlists, you can click one and view the music. What I have is this:
if($result = mysqli_query($link, $playlist))
{
if(mysqli_num_rows($result) > 0)
{
echo "<h1 style='text-align:center;'>Playlists</h1>";
echo "<table style='width:100%'>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . '<a href="getPlaylist.php>' . $row['name']."</td>";
echo "</tr>";
}
echo "</table>";
mysqli_free_result($result);
So I need it so that once someone clicks the link, it will transition them into getPlayList.php and also send over the name in the link, (technically $row['name']) so that I can actually pull up the information in the SQL database. I've tried doing things like:
echo "<td>" . '<a href="getPlaylist.php?link=' . $row["name"]. '">' . $row['name']."</td>";
but to no avail. Can anyone help?
edit:
So what this segment is supposed to do is, it makes a table with each row having a link to a different playlist, but they all go to "getPlaylist.php" where using the name of the link clicked, it will then retrieve all the songs that exist inside it.
edit: solved thanks to RamRaider
I had to use $_GET['link'] in getPlaylist.php
I have this piece of code below. It displays the image and name of all the entries in a table in my database. The name is set up to become a hyperlink.Is it possible to make it so when one specific name is clicked that data for only that specific name will be displayed on the page you are sent to?
So for example if I select the first entry that is displayed back "mealname1" and it takes me to the showrecipe.php page, can I make it so I can display all the data I have for "mealname1" and only "mealname1". I'm really lost, I have scoured the internet and my php books but can't find anything to that is relevant.
If there is no way of doing it is there an obvious solution that I am missing?... I am very much a novice to this... thanks for your help guys.
<?php
require("db.php");
$prodcatsql = "SELECT * FROM recipes";
$prodcatres = mysql_query($prodcatsql);
$numrows = mysql_num_rows($prodcatres);
if($numrows == 0)
{
echo "<h1>No Products</h1>";
echo "There are no recipes available right now.";
}
else
{
echo "<table id='recipetable'>";
while($prodrow = mysql_fetch_assoc($prodcatres))
{
echo "<tr>";
if(empty($prodrow['image'])){
echo "<td><img
src='./images/No_image.png' alt='"
. $prodrow['mealname'] . "'></td>";
}
else {
echo "<td><img src='./images/".$prodrow['image']
. "' alt='"
. $prodrow['mealname'] . "'></td>";
}
echo "<td>";
echo ''.$prodrow['mealname'].'';
echo "</td>";
echo "</tr>";
}
echo "</table>";
}
?>
Change the query to
SELECT * FROM recipes WHERE mealname='$mealname' LIMIT 1;
You can remove "LIMIT 1" if you want but this makes sure you will only get 1 or 0 row back. Don't forget to escape the string.
Sorry if this is a noob question, but I'm still getting up to speed with PHP and can't find an answer to this one.
I have a php script that queries a mySQL table and then builds an HTML table from the results. This all works just fine. As part of that script, I add a <td> to each <tr> that gives the user a chance to delete each specific record from the database, one by one, if they so choose.
To make this work, I have to be able to pass over to the php script the unique identifier of that record, which exists as one of the values. Problem is, I don't know how to pass this value.
Here is my php script that builds the HTML table:
while ($row = mysql_fetch_array($result)) {
echo
"<tr class=\"datarow\">" .
"<td id=\"id_hdr\">" . $row['id'] . "</td>" .
"<td id=\"name_hdr\">" . $row['name'] . "</td>" .
"<td id=\"btn_delete\">
<form action=\"delete_item.php\">
<input type=\"image\" src=\"images/delete.png\">
</form>
</td>" .
"</tr>";
}
So, somehow I either need to explicitly pass 'id' along with "delete_item.php" and/or find a way on the php side to capture this value in a variable. If I can accomplish that I'm home free.
EDIT: Trying to implement both suggestions below, but can't quite get there. Here is how I updated my form based on how I read those suggestions:
"<td id='btn_delete'>".
"<form action='scripts/delete_item.php'>".
"<img src='images/delete.png'>".
"<input type='hidden' id='uid' value='" . $row['id'] . "'>".
"<input type='submit' value='Submit'/>".
"</form>".
"</td>" .
Then, in delete_item.php, I have this:
$id = $_POST['uid'];
$sql = "DELETE FROM myTable WHERE id=$id";
$result = mysql_query($sql);
if (!$result) {
die("<p>Error removing item: " . mysql_error() . "</p>");
}
But when I run it, I get the error:
Error removing item: You have an error in your SQL syntax; check the
manual that corresponds to your MySQL server version for the right
syntax to use near '' at line 1
And one final thing: this approach gives me a button with the word 'submit' directly under my image. I'd prefer not to have this if possible.
Thanks again!
<form action=\"delete_item.php\">
<input type=\"hidden\" value=\"$row['id']\" name=\"uid\" >
<input type=\"image\" src=\"images/delete.png\">
</form>
The unique id is placed in a hidden input. You can get this value using
$_POST['uid']
But you need to submit the form
<input type=\"submit\" name=\"submit\" value=\"delete\" ">
You could use an anchor tag with parameter for id.
ie, www.example.com/delete.php?id=20
Now you could get that id on page delete.php as $_GET['id']
Using that you could delete the data from the table and return to the required page by setting up header
If you required you could use the same logic with AJAX and with out a page reload you could permenently delete that data. I would recommend AJAX