I have a search form on a previous page where I allow users to search for $q. I then query the database for 'keys' LIKE $q. The while loop displays the 'name' and 'weblink' of each matching database entry.
This all works correctly. However, I would like the 'weblink' to display as a clickable link. Ideally it would read as if it were HTML: 'weblink'. I cannot figure out the correct combo of php and html to make both the while loop, and the HTML work.
Any help would be appreciated.
Thanks in advance.
// query database
$query = mysql_query("SELECT * FROM `forumlist` WHERE `keys` LIKE '%$q%'");
// display query results
while($row = mysql_fetch_array($query))
{
echo $row['name'];
echo "<br/>";
echo $row['weblink'];
}
while($row = mysql_fetch_array($query))
{
echo $row['name'];
echo "<br/>";
echo '' . $row['weblink'] . '';
}
Related
I am just running a SQL query to fetch a particular data from the database and would like to display in horizontal way inspite of the vertical format, Can we do something to display the data as required.
For Getting the clear understanding of the question i have attached the Output which i am getting and the output which i required after the SQL query is executed.
Thanks in Advance.
Query:
<?php
include 'connect-db.php';
$query = "SELECT distinct work_component FROM daily_progress_demo WHERE package_no='$package_no'";
$result = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($result)){
echo "<center>".$row['work_component']."</center>";
echo "<br />";
}
?>
you can do like this
echo "<table>"
echo "<tr>"
while($row = mysql_fetch_array($result)){
echo "<td>".$row['work_component']."</td>";
}
echo "</tr>";
echo "</table>";
echo $row['work_component']."<br>";
I really don't know how to explain this question. But, I'm trying to make a product page that shows the same row as it is currently showing (so example "go onto a tech product page and at the bottom it shows all other tech related products").
Here is what I've tried:
<?php
$pType = $row['typeP'];
$sql = "SELECT * FROM products WHERE type=$pType";
$result = mysqli_query($con, $sql) or die(mysqli_error($con));
while ($row = mysqli_fetch_array($result)){
echo "<a href='products.php?id={$row['productID']}'>";
echo "<img src='uploads/{$row['displayIMG']}'>";
echo "</a>";
}
?>
Your error would be in the $typeP declaration. I am guessing $row is not defined before that. therefor it is empty. Maybe you were trying to use a GET?
<?php
$pType = $_GET['typeP']; //This would get the type id from the querystring
$sql = "SELECT * FROM products WHERE type='".$pType."'";
$result = mysqli_query($con, $sql) or die(mysqli_error($con));
while ($row = mysqli_fetch_array($result)){
echo "<a href='products.php?id=".$row['productID']."'>";
echo "<img src='uploads/".$row['displayIMG']."'>";
echo "</a>";
}
?>
I should also mention that your code is subject to SQL injection. Please clean your inputs before running them through you database. Otherwise you are asking for trouble.
http://www.wikihow.com/Prevent-SQL-Injection-in-PHP
I am trying to display results from a database onto my site. I have one entry in the database, and the result of that entry should be a "1". However, when I run the following code, it returns "1111111111111..." and continues to load ones. If there are 4 entries, I want the results to be displayed like:
1
2
3
4
I have looked around on this forum and others but cannot find anywhere where somebody had the same problem, and my code looks similar to the others. What did I do wrong?
functions.php:
$sqlgroup = mysql_query ("SELECT groupid FROM groups");
$grouprow = mysql_fetch_array ($sqlgroup);
Settings.php:
<?php while( ($row = $grouprow))
{
echo "<tr>";
echo "<td>".$row['groupid']."</td>";
echo "</tr>";
} ?>
First of all don't use mysql, use mysqli isntead:
$sqlgroup = mysqli_query ("SELECT groupid FROM groups");
while($row = mysqli_fetch_array($sqlgroup))
{
echo "<tr>";
echo "<td>".$row['groupid']."</td>";
echo "</tr>";
}
I'm trying to make a simple search engine using php & mysql . I know how get mysql result using php. Its Work.
ex : - Mysql database stored this (database table filed "meta")
I am having some difficulty with my PHP/MySQL search. It worked great
a year ago, but for some reason it has just stopped working. I have
tried searching and cannot figure it out. I have it echo my query but
it never adds in the WHERE clause. Any help is much appreciated. Form
Code: PHP Code:
Search word is searching
I Need to dispay search result like this
I have tried searching and cannot figure it out. I have it echo my
query but it never adds in the WHERE clause. Any help is much
appreciated. Form Code: PHP Code:
$query= mysql_query("SELECT meta FROM data WHERE LIKE '%searching%' ");
$count = mysqli_num_rows($query);
if($count >0){
while($row = mysqli_fetch_array($query)) {
echo "$row[id]'>$row[name]</option>";
$count =1;
}
else{
echo "<option value='' selected='selected'></option>";
}
mysqli_close($con);
From the code above here, there's a couple of issues that stand out:
The echo for the option doesn't have an open for , so it's not going to show.
When you're echoing information from the $row variable, you're referencing id and name, but the query is only outputting meta.
The query isn't searching on a field for the where clause, and is therefore failing
The code, from what I can see, should be changed to the following to work (assuming no other surrounding issues):
$query = mysql_query("SELECT meta, id, name FROM data WHERE meta LIKE '%searching%'");
$count = mysqli_num_rows($query);
if ($count > 0)
{
while ($row = $mysqli_fetch_array($query))
{
echo "<option value='" . $row['id'] . "'>" . $row['name'] . "</option>";
$count = 1; // not sure why this needs to be here, but it's in your code so I'm keeping it there
}
}
else
{
echo "<option value='' selected='selected'></option>";
}
mysqli_close($con);
I have made a website in which the users can select a value from a dropdown menu and get some information from a database. I used ajax to send the request to the database (so the page doesn't get refreshed when I send the request). Here is the part of the jquery function:
$.ajax({
type:'POST',
url:'activities_code.php',
data: {datastr:datastr, datastr1:datastr1},
success:function(response){
$("#msg").html(response);
}});}); // there are other functions before..
The results appear on the main container of the webpage. They are composed of a title and some text. I echo the title in such a way so it is a link. I also give to each element an id and a class so I can call it later. Here is the corresponding code:
echo "<table id=\"container\">";
$num_results = 0;
while ($row=mysqli_fetch_array($result, MYSQLI_ASSOC)) {
// Here the columns of title and information are printed
echo "<tr><td>";
echo "".$row['title']."";
echo "<br>";
echo $row['PK'];
echo "</td></tr>";
echo "<tr><td>";
echo $row['Information'];
echo "</td></tr>";
}
What I am trying to do now is: When I click on the title (which is a link), a new page to open in which a php script runs a query and show more information:
Here is what I have:
<?php
include('connect.php');
$query = "SELECT title,Information from activities where title='?????'";
$result = mysqli_query($dbcon, $query) or die('no available data');
echo "<table>";
$num_results = 0;
while ($row=mysqli_fetch_array($result, MYSQLI_ASSOC)) {
// Here the columns of title and information are printed
echo "<tr><td>";
echo "".$row['title']." ";
echo "</td></tr>";
echo "<tr><td>";
echo $row['Information'];
echo "</td></tr>";
// Here I sum up the number of the results
$num_results=$num_results+1;
}
?>
I am trying to find a way to put in my query, in the where clause, the name of the title that I selected:
$query = "SELECT title,Information from activities where title='?????'";
Any help would be much appreciated. Let me know if everything is clear or I didn't explain some point clearly.
Thanks.
D.
You can get the title using $_GET global variable and URL parameter. Try changing this line:
echo "".$row['title']."";
to
echo "".$row['title']."";
then you can get the title with these code:
$title = $_GET['title'];
make sure you sanitize the value first. I hope this will help you.
link:
PHP $_GET