When I retrieve a blob via PHP it only loads about half the image.
Here's my code:
<?php
require 'dbconnect.php';
$q="SELECT * FROM mtgcards WHERE Name LIKE '%".$_POST['search']."%'";
$r = mysqli_query($dbc, $q);
while ($row = mysqli_fetch_array($r)) {
echo '<br>'. 'Name: ' . $row['Name'] . ' Mana cost: ' . $row['Mana Cost'] . ' Colour: ' . $row['Colour'] . ' Set: ' . $row['Set'] . ' Ability: ' . $row['Ability']. '<img src="data:image/jpeg;base64,' . base64_encode($row['Image']) . '" width="223" height="311">';
}
mysqli_close($dbc);
?>
Here's what the full picture is and what the picture when I retrieve it comes out as.
Thanks in advance!
As it seems and was determined in comments, BLOB is too small a column type for the image's (raw) size, therefore you will need to use LONGBLOB.
Reference:
http://dev.mysql.com/doc/refman/5.7/en/blob.html
So you will need to alter your column, and resave your data then start over.
I am getting no where with this, I am not getting any output from my echo ,can someone help, thanks in advance...singhy
code below...
$strSQL = "SELECT * FROM <tablename> WHERE id='" . $_GET["serviceName"] . "'";
$rs = mysql_query($strSQL);
while($row = mysql_fetch_array($rs)){
echo "<dt>Name:</dt><dd>" . $row["serviceType"] . " " . $row["serviceName"] . "</dd>";
echo "<dt>Phone:</dt><dd>" . $row["Phone"] . "</dd>";
echo "<dt>Birthdate:</dt><dd>" . $row["BirthDate"] . "</dd>";
}
// Close the database connection
mysql_close();
?>
<p>Return to the list</p>
</body>
</html>
can someone tell me where i am going wrong, ive tried various options, thanks in advance, singhy
Try this debugging code:
$serviceName = mysql_real_escape_string($_GET['serviceName']); // Read PS note at the end
$strSQL = "SELECT * FROM `tablename` WHERE id='$serviceName'";
$rs = mysql_query($strSQL) or die(mysql_error()); // Display any query error
echo "Total number of rows: ". mysql_num_rows($rs); // Echo number of rows
while($row = mysql_fetch_assoc($rs)){
echo "<dt>Name:</dt><dd>" . $row["serviceType"] . " " . $row["serviceName"] . "</dd>";
echo "<dt>Phone:</dt><dd>" . $row["Phone"] . "</dd>";
echo "<dt>Birthdate:</dt><dd>" . $row["BirthDate"] . "</dd>";
}
Please note
You should escape the $_GET request and never use it directly in a query statement. Use mysql_real_escape_string() for that. (This method will be deprecated, read next bullet)
many of the functions you are using will be deprecated starting php 5.5.0 Alternatively you can use PDO prepared statements
replace
$_GET["serviceName"]
with this
$_GET['serviceName']
use single quotes in $_GET in your case.
I tried to use mysqli in for my forum database. this is the code I used:
<meta charset="utf-8">
<?php
include("config.php");
$limits = "6";
$forum_id = "2";
$db = new mysqli($INFO['sql_host'], $INFO['sql_user'], $INFO['sql_pass'], $INFO['sql_database']);
$topics = $db->query("
SELECT
`topics`.`start_date`,
`topics`.`title`,
`topics`.`starter_name`,
`topics`.`posts`,
`topics`.`title_seo`,
`topics`.`tid`,
`posts`.`post`
FROM
`" . $INFO['sql_tbl_prefix'] . "topics` as `topic`,
`" . $INFO['sql_tbl_prefix'] . "posts` as `post`
WHERE
`topics`.`approved` = 1 AND
`topics`.`forum_id`= " . $forum_id . " AND
`posts`.`topic_id` = `topic`.`tid` AND
`posts`.`new_topic` = 1
ORDER BY
`topics`.`start_date`
DESC LIMIT 5");
echo '<ul id="news">';
while ($topic = $topics->fetch_object()) {
$url = $INFO['board_url'] . '/index.php?/topic/' . $topic->tid . '-' . $topic->title_seo . '/';
$topic->post = strip_tags(str_replace(array('[', ']'), array('<', '>'), $topic->post));
$topic->start_date = date("Y.m.d H:i", $topic->start_date);
echo '
<div class="news">
<div class="newsp"><div class="pteksts">' . $topic->title . '</div></div>
<center><img src="img/news.png"></center>
<div class="teksts" style="padding-bottom: 5px;">' . $topic->post . '</div>
</div>
';
}
echo '</ul>';
?>
and errors i received:
Fatal error: Call to a member function fetch_object() on a non-object in /home/public_html/scripts/news.php on line 35
You give aliases for your tables as topic and post, but then you use the aliases topics and posts. You need to change the table qualifiers to use the same spelling as your table alias.
Wrong, because alias topic is not the same as table qualifier topics:
SELECT
`topics`.`start_date`, . . .
FROM
`" . $INFO['sql_tbl_prefix'] . "topics` as `topic`,
. . .
Right, after changing the table qualifier to match the alias name:
SELECT
`topic`.`start_date`, . . .
FROM
`" . $INFO['sql_tbl_prefix'] . "topics` as `topic`,
. . .
Right as well, but alias is unnecessary if it's the same as the base table name:
SELECT
`topics`.`start_date`, . . .
FROM
`" . $INFO['sql_tbl_prefix'] . "topics` as `topics`,
. . .
But more to the point, you should always check the return value from $db->query(), because it returns false if there's an error. You can't call any method on a false because that's not an object.
If that happens, report the error but do not try to fetch from the result. It won't work.
$topics = $db->query(...);
if ($topics === false) {
die($db->error);
}
// now we can be sure it's safe to call methods on $topics
while ($topic = $topics->fetch_object()) {
. . .
Re your comment that the output is blank:
I just tested this script and it mostly works, so I can't guess what's going wrong. I suggest you read your http server's error log, which is where many PHP notices and errors are output.
I do see the following notice:
Notice: A non well formed numeric value encountered in /Users/billkarwin/workspace/SQL/22159646.php on line 51
The line is this:
$topic->start_date = date("Y.m.d H:i", $topic->start_date);
The problem is that PHP's date() function takes an integer timestamp, not a date string.
You might want to format the date in SQL, using MySQL DATE_FORMAT() function instead.
I am having a small issue with some coding of mine. For some reason my entries aren't dropping in my DB. Any suggestions would be greatly appreciated! Here is my code...
<?php
$dbhost="localhost";
$dbname="DBNAME";
$dbuser="USER";
$dbpasswd="PASSWORD"; // connect to the db
$dbcxn = mysqli_connect($dbhost, $dbuser, $dbpasswd);
if (!$dbcxn) {
die('Could not connect: ' . mysql_error());
}
$db_selected = mysqli_select_db($dbcxn, $dbname);
if (!$db_selected) {
die ('Can\'t use dbreviews : ' . mysql_error());
}
$query = "INSERT INTO entries ( submitterFirstName, submitterLastName, submitterPhone, submitterEmail, referredFirstName, referredLastName, referredPhone, referredEmail, referredReason)
VALUES ('$submitterFirstName', '$submitterLastName', '$submitterPhone', '$submitterEmail', '$referredFirstName', '$referredLastName', '$referredPhone', '$referredEmail', '$referredProject')";
$result=mysqli_query($dbcxn, $query);
?>
The first thing you want to check is echo the query back to yourself and read it over.
Second, check the table structure. Make sure the column names are all spelled correctly and that all fields exist in your table (I've accidently forgotten to add a column before).
Third, you may or may not receive error messages depending on your configuration. But, you can manually check.
if (!$result) {
echo mysqli_error($dbcxn);
}
First thing first should be code formatting, it will help you read the code and consequently find your errors easier.
$query = "
INSERT INTO
entries
(
submitterFirstName,
submitterLastName,
submitterPhone,
submitterEmail,
referredFirstName,
" .
"referredLastName,
referredPhone,
referredEmail,
referredReason
)
" .
" VALUES
(
'$submitterFirstName',
'$submitterLastName',
'$submitterPhone',
' $submitterEmail',
'$referredFirstName'," .
"'$referredLastName',
'$referredPhone',
'$referredEmail',
'$referredProject'
);
"
The above is your query string split onto several lines, there are some errors which should be evident straight away? Once formatted I would do echo $query and view the output of $query.
Also try seeing if you can do an insert without using php (using mysql workbench, php admin etc) then compare it with the string value you have set as $query.
// less errors, please note that inside "" you can include php $vars without needing to escape.
$query = "
INSERT INTO
entries
(
submitterFirstName,
submitterLastName,
submitterPhone,
submitterEmail,
referredFirstName,
referredLastName,
referredPhone,
referredEmail,
referredReason
)
VALUES
(
'$submitterFirstName',
'$submitterLastName',
'$submitterPhone',
'$submitterEmail',
'$referredFirstName',
'$referredLastName',
'$referredPhone',
'$referredEmail',
'$referredProject'
);
";
Change your query variable to:
$query = "INSERT INTO entries " .
"( submitterFirstName, submitterLastName, submitterPhone, submitterEmail, referredFirstName, " .
" referredLastName, referredPhone, referredEmail, referredReason )" .
" VALUES ('" .
$submitterFirstName . "', '" .
$submitterLastName . "', '" .
$submitterPhone . "', '" .
$submitterEmail . "', '" .
$referredFirstName . "', '" .
$referredLastName . "', '" .
$referredPhone . "', '" .
$referredEmail . "', '" .
$referredProject . "')";
and it should be working.
Suggesting to use mysqli prepare
Im starting to learn PHP. When I run the script it had an error that said: "Assigned Employee:resource(6) of type (mysql result)" . Please help me and sorry for my bad English Here is the code:
include_once 'rnheader.php';
include_once 'rnfunctions.php';
</tr><tr><td><label for="AssignedEmp"> Assigned Employee:</label></td><td>';
$query = "SELECT UserName FROM employee where Classification_ClassificationID = '2'";
$result = queryMysql($query);
if (!queryMysql($query)) {
echo "Query fail: $query<br />" .
mysql_error() . "<br /><br />";
}
else
{
var_dump($result);
exit;
<select name = "UserName", "Name" size = "1">'; // or name="toinsert[]"
while ($row = mysqli_fetch_array($result)) {
'<option value="' . htmlspecialchars($row['UserName']) . '" >'
. htmlspecialchars($row['UserName'])
. '</option>';
}
}
'</select>';
?>
I isn't error. That output is produced by this line:
var_dump($result);
exit;
Since you are dumping the result of a query directly it is dumping a resource object and then you are immediately exiting the application. See after a query, you get data in a resource object, which is why we use the while loop that you have later. Remove the
exit;
And see what you get after you see
resource(6) of type (mysql result)
What is the queryMysql function that you have build? Can we see that?
Also, you have quotes here:
Classification_ClassificationID = '2'
Quotes are for strings, varchars, blobs, etc. An ID is typically an integer. Is your Classification_ClassificationID a varchar in your database or an integer. If it is an integer, take out the single quotes.