Display SQL query results

Question

I am having trouble displaying results from a SQL query. I am trying to display all images and prices from a products table.

I am able to display the echo statement "Query works" in the browser. But, the results are not displaying in the browser.

        if ($count > 0) {
            echo "Query works";
        } else {
            echo "Query doesn't work" ."<br/>";
        }

PHP Code:

$con = getConnection();
        $sqlQuery = "SELECT * from Products";

        // Execute Query -----------------------------           
        $result = mysqli_query($con, $sqlQuery);
            if(!$result) {
                echo "Cannot do query" . "<br/>";
                exit;
            }

            $row = mysqli_fetch_row($result);
            $count = $row[0];

            if ($count > 0) {
                echo "Query works";
            } else {
                echo "Query doesn't work" ."<br/>";
            }

          // Display Results -----------------------------

            $num_results = $result->numRows();

            for ($i=0; $i<$num_results; $i++) {
                $row = $result->fetchRow(MDB2_FETCH_ASSOC);
                echo '<img src="'.$row['Image'].'>';
                echo "<br/>" . "Price: " . stripslashes($row['Price']);

}

Screenshot 1
alt text Screenshot 2: removed the images from the database, and used a filepath instead

alt text Screenshot 3: print_r($row)

alt text

Did you try to print the rows you got with print_r($row)

svk
– svk

2010-09-20 07:24:07 +00:00
Commented Sep 20, 2010 at 7:24 — svk
– svk, Commented Sep 20, 2010 at 7:24

Nik · Accepted Answer · 2010-09-20 07:30:13Z

2

try

$sqlQuery = "SELECT * from Products";

        // Execute Query -----------------------------           
        $result = mysqli_query($con, $sqlQuery);
            if(!$result) {
                echo "Cannot do query" . "<br/>";
                exit;
            }

            $row = mysqli_fetch_row($result);
            $count = $row[0];

            if ($count > 0) {
                echo "Query works";
            } else {
                echo "Query doesn't work" ."<br/>";
            }

          // Display Results -----------------------------

            $num_results =mysqli_num_rows($result);

            for ($i=0; $i<$num_results; $i++) {
                $row = mysqli_fetch_assoc ($result);
                //print_r($row);
              echo '<img src="'.$row['Image'].'>';
                echo "<br/>" . "Price: " . stripslashes($row['Price']);
            }

answered Sep 20, 2010 at 7:30

Nik

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6 Comments

jc70 Over a year ago

@Yogesh i've tried the various solutions suggested, and with yours i was able to display something in the browser--which is a good sign. but right now, the images are showing up as characters. i tried removing the images from the database, and just putting the file path, but in both cases the images display as characters.

Nik Over a year ago

Can you print the row and check whether image path is coming properly or not.

jc70 Over a year ago

Array ( [ProductID] => 2 [ProductDescription] => Atheletic Description. [ProductType] => Athletic [Image] => images/shoe_box.jpg [Price] => 100.00 )

jc70 Over a year ago

I took a screenshot of what is displayed on my browser when I did print_r($row). The comment above, is only one of the records, the rest are in the screenshot.

jc70 Over a year ago

Also noticed that only 4 out of the 9 records in the database are displaying.

|

codaddict · Accepted Answer · 2010-09-20 07:22:28Z

2

I think

$row = mysqli_fetch_row($result);
$count = $row[0];

should be

$count = $result->numRows();
if ($count > 0) {
  echo "Query produced $count rows";
} else {
  echo "Query produced no rows" ."<br/>";
  return;
}

And your for loop should use fetch_assoc as:

while ($row = $result->fetch_assoc()) {
  echo '<img src="'.$row['Image'].'>';
  echo "<br/>" . "Price: " . stripslashes($row['Price']);
}

answered Sep 20, 2010 at 7:22

codaddict

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Comments

Jochem · Accepted Answer · 2010-09-20 07:20:54Z

1

$row is the first result-row (if any) from your query. $row[0] is the first column in this query (which, since you use select *, depends on the order of the columns in your database). So, whether $row[0] > 0 depends on the content of your database.

answered Sep 20, 2010 at 7:20

Jochem

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Comments

Alex · Accepted Answer · 2010-09-20 08:43:10Z

1

It's displaying characters because that is how you have stored the image. In order to show the image you are going to have to draw the image with something like:

echo '<img src="data:image/gif;base64,'.base64_encode($row['Image']).'" />';

answered Sep 20, 2010 at 8:43

Alex

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Comments

Fanis Hatzidakis · Accepted Answer · 2010-09-20 07:22:30Z

0

Mysqli doesn't have fetchRow(), that's part of the Pear::MDB2 library

See the docs: http://www.php.net/manual/en/mysqli-result.fetch-assoc.php

Change your loop to the following:

while ($row = $result->fetch_assoc()) {
    echo '<img src="'.$row['Image'].'>';
    echo "<br/>" . "Price: " . stripslashes($row['Price']);
}

Also, by doing this:

$row = mysqli_fetch_row($result);
$count = $row[0];

before the loop you are essentially skipping the first row and not displaying its image in the loop.

answered Sep 20, 2010 at 7:22

Fanis Hatzidakis

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Comments

In81Vad0 · Accepted Answer · 2010-09-24 11:09:19Z

0

to print all results from a query you can use a while loop

while($row=mysqli_fetch_assoc($result)){
    echo 'Price '.$row['Price'].'<br/>';
}

edited Sep 24, 2010 at 11:09

answered Sep 20, 2010 at 7:28

In81Vad0

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Comments

user1932459 · Accepted Answer · 2013-02-23 20:02:48Z

0

Instead of

If ($count > 1)

Try

If ($count >= 1)

answered Feb 23, 2013 at 20:02

user1932459

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