Below is my code. It is working correctly
but the image is not displayed on my web page. How to rectify this?
while($row = mysqli_fetch_assoc($result))
{
echo "id: " . $row["product_id"]. " - Name: " .$row["product_name"]. " . $row["product_type"]. "Price " .$row["price"]. "photo" .$row["product_photo"]. "<br>";
$photo=$row["product_photo"];
echo '.$photo.';
echo '<img src='.$photo.'/>';
}
Try the below.
Make sure the full image path is coming up.
<?php
while($row = mysqli_fetch_assoc($result))
{
echo "id: " . $row["product_id"]. " - Name: " . $row["product_name"] . $row["product_type"]. "Price " .$row["price"]. "photo" .$row["product_photo"]. "<br>";
$photo=$row["product_photo"];
echo '.$photo.';
echo "<img src='$photo'/>";
}
?>
Related
Sorry for the confusing title. I am as confused.
SO what I am trying to do is print results from the MySQL database I've got. I have a checkbox value as "yes" in my DB and I would like to replace this to some other word while printing out the results.
I've tried different ways but all of them break the page, because I'm new to this and have no idea what I am doing.
Here is my code so far (only put what I think is relevant):
$keyword= "";
if (isset($_POST["keyword"])) {
$keyword = ($_POST["keyword"]);
}
$results = mysqli_query($con, "SELECT * FROM pcdata WHERE name LIKE '$keyword' LIMIT 0, 25");
if (!$results) {
echo "Not found...";
} else {
echo "Found...<br>";
}
while ($row = mysqli_fetch_array($results)) {
echo "<br>";
echo "Name: " . $row['name'] . "<br>";
echo "Model: " . $row['model'] . "<br>";
echo "Operating system: " . $row['model'] . "<br>";
echo "Type of computer: " . $row['pctype'] . "<br>";
echo "Other information: " . $row['info'] . "<br>";
echo "Need help ASAP: " . $row['help'] . "<br>";
}
Why don't you try a simple if inside your while:
$myvariable='';
if($row['help']='yes'){
$myvariable='put_something_here';
}
And in your echo just do:
echo "Need help ASAP: " . $myvariable . "<br>";
Or a ternary solution:
$row['help'] == 'yes' ? 'put_something_here' : 'what_do_you_want_to_print_if_it_is_not_yes'
Try this code:
$keyword= "";
if (isset($_POST["keyword"]))
$keyword=($_POST["keyword"]);
$results=mysqli_query($con,"
SELECT *
FROM pcdata
WHERE name LIKE '$keyword' LIMIT 0,25");
if (!$results) {
echo "Not found...";
}
else {
echo "Found...<br>";
}
while ($row = mysqli_fetch_array($results))
{
echo "<br>";
echo "Name: " . $row['name'] . "<br>";
echo "Model: " . $row['model'] . "<br>";
echo "Operating system: " . $row['model'] . "<br>";
echo "Type of computer: " . $row['pctype'] . "<br>";
echo "Other information: " . $row['info'] . "<br>";
echo "Need help ASAP: ";
if ($row['help'] === 'yes'){
echo 'YES';
} else {
echo 'NO';
}
echo '<br>';
}
We check the value of $row['help'] and if it "yes" printing 'YES', if other - printing 'NO'
you can also use select statement combined with Case statement which will result as desired try following code
$keyword= "";
if (isset($_POST["keyword"]))
{
$keyword = ($_POST["keyword"]);
}
//used different variable to build query
$selectquery="SELECT id,name,model,pctype,info CASE WHEN help='yes' THEN 'Your Yes String' WHEN help='no' THEN 'Your No String' else 'nothing' END as help FROM pcdata where name like '$keyword'" ;
//passed $selectquery to mysqli_qery
$results = mysqli_query($con, $selectquery);
if (!$results) {
echo "Not found...";
} else {
echo "Found...<br>";
}
while ($row = mysqli_fetch_array($results)) {
echo "<br>";
echo "Name: " . $row['name'] . "<br>";
echo "Model: " . $row['model'] . "<br>";
echo "Operating system: " . $row['model'] . "<br>";
echo "Type of computer: " . $row['pctype'] . "<br>";
echo "Other information: " . $row['info'] . "<br>";
echo "Need help ASAP: " . $row['help'] . "<br>";
}
<?php
$id = '2422414574';
$json = file_get_contents("http://xxxxxx.com/api.php?token=xxxx&id=xxxx");
$data = json_decode($json);
echo $data->phim[0]->filmName . "<br/>";
echo $data->phim[0]->epsList[0]->name . " - ";
echo $data->phim[0]->epsList[0]->id . "<br/>";
echo $data->phim[0]->epsList[1]->name . " - ";
echo $data->phim[0]->epsList[1]->id . "<br/>";
echo $data->phim[0]->epsList[2]->name . " - ";
echo $data->phim[0]->epsList[2]->id . "<br/>";
echo $data->phim[0]->epsList[3]->name . " - ";
echo $data->phim[0]->epsList[3]->id . "<br/>";
echo $data->phim[0]->epsList[4]->name . " - ";
echo $data->phim[0]->epsList[4]->id . "<br/>";
echo $data->phim[0]->epsList[5]->name . " - ";
echo $data->phim[0]->epsList[5]->id . "<br/>";
echo $data->phim[0]->epsList[6]->name . " - ";
echo $data->phim[0]->epsList[6]->id . "<br/>";
echo $data->phim[0]->epsList[7]->name . " - ";
echo $data->phim[0]->epsList[7]->id . "<br/>";
echo $data->phim[0]->epsList[xxxx]->name . " - ";
echo $data->phim[0]->epsList[xxxx]->id . "<br/>";
echo xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
echo xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
As you can see, I have to repeat 1 code in many time. Is there any short way to do this without repeat code?
For example, I want to get all values from id and name from this URL, which have 24 items. That means I have to repeat code in 24 times :(
what about foreach()?
foreach($data->phim as $phim)
{
echo $phim->filmName . "<br/>";
foreach($phim->epsList as $epsList)
{
echo $epsList->name . " - ";
echo $epsList->id . "<br/>";
}
}
You need to iterate over it using a foreach loop. This is one of the basic functions of PHP and there are many tutorials for it.
I have this PHP code, which fetches data from my SQL database called "comments".
This code prints out every comment in the table:
<?php
$sql = "SELECT id,name,email,number,text FROM comments";
$result = $conn->query($sql);
if($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "<strong>ID:</strong><br> " . $row["id"] . "<br>";
echo "<strong>Navn:</strong><br> " . $row["name"] . "<br>";
echo "<strong>Email:</strong><br> " . $row["email"] . "<br>";
echo "<strong>Nummer:</strong><br> " . $row["number"] . "<br>";
echo "<strong>Melding:</strong><br> " . $row["text"] . "<br><br><br>";
}
echo '<div class = "white_line_comments"></div>';
} else {
echo "0 results";
}
This has worked fine so far, everything prints as it's supposed to.
Then I decided I wanted a way to give each individual comment some sort of identification to make them unique. I tried putting each single comment into its own div, using the SQLtable row id as id for the div.
However, when I try to access my webpage now, it tells me the website doesn't work (HTTP Error 500).
<?php
$sql = "SELECT id,name,email,number,text FROM comments";
$result = $conn->query($sql);
if($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "
<div class='$row[' id'].'>";
echo "<strong>ID:</strong><br> " . $row["id"] . "<br>";
echo "<strong>Navn:</strong><br> " . $row["name"] . "<br>";
echo "<strong>Email:</strong><br> " . $row["email"] . "<br>";
echo "<strong>Nummer:</strong><br> " . $row["number"] . "<br>";
echo "<strong>Melding:</strong><br> " . $row["text"] . "<br><br><br>";
echo '</div>';
}
echo '
<div class="white_line_comments"></div>';
} else {
echo "0 results";
}
Any ideas on this? I guess I must've done something wrong when including the div, but I can't figure out what!
You have an error in this line after starting while loop:
echo "<div class ='$row['id'].'>";
It should be
echo "<div class ='". $row['id'] ."'>";
You should also configure your web server/hosting/localhost to throw a PHP error.
Read this if you are on localhost or your own server:
How can I make PHP display the error instead of giving me 500 Internal Server Error
Read this if you are using shared hosting: How do I displaying details of a PHP internal server error?
echo "<div class ='$row['id'].'>";
First line in while loop should look like this
echo "<div class = '".$row['id']."'>";
or
echo "<div class = '$row['id']'>"
You have mixed different apostrophe, try this:
echo "<div class ='" . $row['id'] . "'>";
I really don't know what I'm doing here. I can display the url from the database but I can't figure out how to add the html code
$sql = "SELECT IMG_URL, Birthdate, FirstName, LastName FROM Student";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "IMG: " . $row["IMG_URL"]. " Birthdate: " . $row["Birthdate"]. " - Name: " . $row["FirstName"]. " " . $row["LastName"]. "<br>";
}
Right now, you're not outputting any HTML at all from this code. Insert an <img> tag around the outputted image:
echo "IMG: <img src=\"" . $row["IMG_URL"]. "\" /> Birthdate: " . $row["Birthdate"]. " - Name: " . $row["FirstName"]. " " . $row["LastName"]. "<br>";
I am performing a routine to check if books exist in a database given a range. I want to echo if no books are found. I have the following:
$search = mysqli_query($con,$query);
while(list($book_id, $title, $authors, $description, $price) = mysqli_fetch_row($search)){
if(!empty($book_id)) {
echo "Book ID: " . $book_id . "<br/>";
echo "Book Title: " . $title . "<br/>";
echo "Authors: " . $authors . "<br/>";
echo "Description: " . $description . "<br/>";
echo "Price: " . $price . "<br/>";
echo "<br/>";
}
if(empty($book_id)){
echo "Fail";
}
}
If no books are found nothing is printed. The echo does not work? How come?
Thanks
Because if no records are returned, you won't enter in the while at all, as $book_is will contain the value false and while(false)... you know
In this situation you may use mysqli_num_rows to check if there are rows found
Your no results echo is inside the while loop which will never get called
while (list($book_id, $title, $authors, $description, $price) = mysqli_fetch_row($search)) {
if (!empty($book_id)) {
echo "Book ID: " . $book_id . "<br/>";
echo "Book Title: " . $title . "<br/>";
echo "Authors: " . $authors . "<br/>";
echo "Description: " . $description . "<br/>";
echo "Price: " . $price . "<br/>";
echo "<br/>";
}
}
if (empty($book_id)) {
echo "Fail";
}
Move it outside like the code above
It is because your while loop has not been accessed at all. Try to echo something out of the if conditions, and see if it get's displayed.