what is error in this code at mysql query???
i think it is very easy question but error could not found :( while this mysql query is working in SQLYog
$result =mysql_query(SELECT NUMPOINTS((i.geom))COUNT FROM district i WHERE district_name="D.G.KHAN";)
compete code is
//$type="R";
$result =mysql_query(SELECT NUMPOINTS((i.geom))COUNT FROM district i WHERE district_name="D.G.KHAN";)
$row2 = mysql_fetch_array($result);
$count=$row2['count'];
$x_points = array();
$y_points = array();
/* $row1 ['color']="00000"; */
for($c=1;$c<=$count;$c++){
$res=mysql_query("SELECT district_name,X ( POINTN ( i.geom ,$c))xx ,Y( POINTN ( i.geom ,$c))yy ,color FROM district i where dis_id='".$id."'" );
while($row1 = mysql_fetch_array($res)){
array_push($x_points['xx']);
array_push($y_points['yy']);
}
}
print_r ($x_points);
print_r ($y_points);
?>
You need to add as a string:
$result =mysql_query("SELECT NUMPOINTS((i.geom)) FROM district i WHERE district_name='D.G.KHAN'";)
Related
I have this in my PHP.
$sql="select * from defaulttime";
$result = mysqli_query($db, $sql);
while ($row = mysqli_fetch_array($result)){
$row_set[] = $row;
}
echo trim(json_encode($row_set));
In my SQL some columns are Integer. but in JSON it shows as String.
can someone please tell me what I am doing wrong?
when I run the PHP, I get the below results. (Note Hour and Minute should be Integers)
[{"name":"Test","Hour": "6" ,"Minute":"45"}],
I want it to show string as string and Int as Int.
{"name":"Test","Hour": 6 ,"Minute":45}
You have to use JSON_NUMERIC_CHECK option in order to get numerics value,
look here: https://lornajane.net/posts/2011/php-returning-numeric-values-in-json
echo trim(json_encode($row_set,JSON_NUMERIC_CHECK));
Thanks all,
I got it working by using this code:
$result = mysqli_query($db, $sql);
while ($row = mysqli_fetch_array($result)){
$row_set[] = $row;
}
$json=json_encode($row_set,JSON_NUMERIC_CHECK);
echo $json;
Recently I was making an upload component for joomla 3.1 back-end.
based on How to Save Uploaded File's Name on Database
I was successful in moving the file to the hard-drive,
however I just cant get the update query to work based on the posted post above.
I don't get any SQL errors and saving works, but somehow ignores the database part.
I really hope I missed something obvious. (btw I don't know the joomla way of queries very well)
In phpmyadmin the following query works:
UPDATE hmdq7_mysites_projects
SET project_file = 'test'
WHERE id IN (3);
I have tried the following queries:
$id = JRequest::getVar('id');
$db =& JFactory::getDBO();
$sql = "UPDATE hmdq7_mysites_projects
SET project_file =' " . $filename. "'
WHERE id IN (".$id.");";
$db->setQuery($sql);
$db->query();
$colum = "project_file";
$id = JRequest::getVar('id');
$data = JRequest::getVar( 'jform', null, 'post', 'array' );
$data['project_file'] = strtolower( $file['name']['project_file'] );
$db =& JFactory::getDBO();
$query = $db->getQuery(true);
$query->update('#__mysites_projects');
$query->set($column.' = '.$db->quote($data));
$query->where('id'.'='.$db->quote($id));
$db->setQuery($query);
$db->query();
Here is the current code:
class MysitesControllerProject extends JControllerForm
{
function __construct() {
$this->view_list = 'projects';
parent::__construct();
}
function save(){
// ---------------------------- Uploading the file ---------------------
// Neccesary libraries and variables
jimport( 'joomla.filesystem.folder' );
jimport('joomla.filesystem.file');
$path= JPATH_SITE . DS . "images";
// Create the gonewsleter folder if not exists in images folder
if ( !JFolder::exists(JPATH_SITE . "/images" ) ) {
JFactory::getApplication()->enqueueMessage( $path , 'blue');
}
// Get the file data array from the request.
$file = JRequest::getVar( 'jform', null, 'files', 'array' );
// Make the file name safe.
$filename = JFile::makeSafe($file['name']['project_file']);
// Move the uploaded file into a permanent location.
if ( $filename != '' ) {
// Make sure that the full file path is safe.
$filepath = JPath::clean( JPATH_SITE . "/images/" . $filename );
// Move the uploaded file.
JFile::upload( $file['tmp_name']['project_file'], $filepath );
$colum = "project_file";
$id = JRequest::getVar('id');
$data = JRequest::getVar( 'jform', null, 'post', 'array' );
$data['project_file'] = strtolower( $file['name']['project_file'] );
$db =& JFactory::getDBO();
$query = $db->getQuery(true);
$query->update('#__mysites_projects');
$query->set($column.' = '.$db->quote($data));
$query->where('id'.'='.$db->quote($id));
$db->setQuery($query);
$db->query();
}
// ---------------------------- File Upload Ends ------------------------
JRequest::setVar('jform', $data );
return parent::save();
}
(Answered by the OP in comments. Converted to a community wiki answer. See Question with no answers, but issue solved in the comments (or extended in chat) )
The OP wrote:
Solved: after reviewing the post update record in database using jdatabase I made up some fixed test values. It turns out the query is correct but $data variable in the query had no data. $data['project_file'] = strtolower( $file['name']['project_file'] ); removed the array from first part and variable worked.
I am trying to export my MySQL tables from my database to a JSON file, so I can list them in an array.
I can create files with this code no problem:
$sql=mysql_query("select * from food_breakfast");
while($row=mysql_fetch_assoc($sql))
{
$ID=$row['ID'];
$Consumption=$row['Consumption'];
$Subline=$row['Subline'];
$Price=$row['Price'];
$visible=$row['visible'];
$posts[] = array('ID'=> $ID, 'Consumption'=> $Consumption, 'Subline'=> $Subline, 'Price'=> $Price, 'visible'=> $visible);
}
$response['posts'] = $posts;
$fp = fopen('results.json', 'w');
fwrite($fp, json_encode($response));
fclose($fp);
Now this reads a table and draws it's info from the fields inside it.
I would like to know if it is possible to make a JSON file with the names of the tables, so one level higher in the hierarchy.
I have part of the code:
$showtablequery = "
SHOW TABLES
FROM
[database]
LIKE
'%food_%'
";
$sql=mysql_query($showtablequery);
while($row=mysql_fetch_array($sql))
{
$tablename = $row[0];
$posts[] = array('tablename'=> $tablename);
}
$response['posts'] = $posts;
But now i am stuck in the last line where is says: $ID=$row['ID']; This relates to the info inside the Table and I do not know what to put here.
Also as you can see, I need to filter the Tables to only list the tables starting with food_ and drinks_
Any help is greatly appreciated:-)
There is no 'table id' in MySQL and therefore the result set from SHOW TABLES has no index id. The only index in the resultset is named 'Tables_in_DATABASENAME'.
Also you should use the mysqli library as the good old mysql library is depreacted. Having prepared an example:
<?php
$mysqli = new mysqli(
'yourserver',
'yourusername',
'yourpassword',
'yourdatabasename'
);
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") "
. $mysqli->connect_error;
}
$result = $mysqli->query('SHOW TABLES FROM `yourdatabasename` LIKE \'%food_%\'');
if(!$result) {
die('Database error: ' . $mysqli->error);
}
$posts = array();
// use fetch_array instead of fetch_assoc as the column
while($row = $result->fetch_array()) {
$tablename = $row[0];
$posts []= array (
'tablename' => $tablename
);
}
var_dump($posts);
I have wp_places custom table and I am getting this when I am printing array:
[0] => stdClass Object
(
[home_location] => 24
)
[1] => stdClass Object
(
[home_location] => 29
)
Now I want to implode value like this way (24,29) but in my code I am getting this error:
<b>Warning</b>: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
My Code
$getGroupType = $_POST['parent_category'];
$result = $wpdb->get_results( "SELECT home_location FROM wp_places WHERE blood_group LIKE '".$getGroupType."%'" );
$bgroup = Array();
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
$bgroup[] = implode(',',$row);
}
echo implode(',',$bgroup);
Any ideas or suggestions? Thanks.
$wpdb->get_results() already do the fetching for you, you don't need to call mysql_fetch_array
Given what you want to do, your code should look like this :
$getGroupType = $_POST['parent_category'];
$result = $wpdb->get_results( "SELECT home_location FROM wp_places WHERE blood_group LIKE '".$getGroupType."%'" );
$bgroup = Array();
foreach ($result as $location) {
$bgroup[] = $location->home_location;
}
echo '('.implode(',',$bgroup).')';
It's an PHP object that contains results, it isn't a MySQL Result.
Looking at the docs, it should be used like
foreach ($result as $row) {
$bgroup[] = $row->home_location;
}
echo implode(',',$bgroup)
Please help to create a query!
This is a working example of my query Retrieve data as JSON using PHP:
<?php
$con = mysql_connect("localhost","user","password");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("admin_accounting", $con);
$result = mysql_query("SELECT unix_timestamp(date), sum(ksi2k) FROM accounting where lhc_vo like 'ops' group by year(date), month(date)");
$rows = array();
$rows['type'] = 'area';
$rows['name'] = 'Ops';
while($r = mysql_fetch_array($result)) {
$rows['data'][] = $r[0]*1000;
$rows['data'][] = $r[1];
array_push($rows);
}
print json_encode($rows, JSON_NUMERIC_CHECK);
mysql_close($con);
?>
The JSON results look like this:
{"type":"area","name":"Ops","data":[1167664515000,0,1170342915000,0,1172762115000,0,1175436915000,0,1178028915000,0]}
But I need the JSON results should look like this:
{"type":"area","name":"Ops","data":[[1167664515000,0],[1170342915000,0],[1172762115000,0],[1175436915000,0],[1178028915000,0]]}
I would be very grateful for the help
while($r = mysql_fetch_array($result)) {
$rows['data'][] = array($r[0]*1000, $r[1]);
}
In addition to my other answer, you ought to consider switching away from the ancient (and now deprecated, as of PHP v5.5) ext/mysql. Here is an example using PDO, in which you can see how simple your problem becomes:
<?php
$con = new PDO(
'mysql:hostname=localhost;dbname=admin_accounting',
'user',
'password'
);
$result = $con->query('
SELECT 1000*UNIX_TIMESTAMP(date), SUM(ksi2k)
FROM accounting
WHERE lhc_vo LIKE "ops"
GROUP BY YEAR(date), MONTH(date)
');
print json_encode([
'type' => 'area',
'name' => 'Ops',
'data' => $result->fetchAll(PDO::FETCH_NUM)
]);
?>
Note that:
I have moved the multiplication into the database layer in order that I can simply call fetchAll() to obtain the resulting array;
MySQL will select an indeterminate value from amongst those in each group for the first column in the resultset; should this be undesirable, you will need to apply a suitable aggregate function to the reference to the date column; and
I have used the short array syntax, which is only available from PHP v5.4—if you're using an earlier version, you will need to replace the [ … ] of the argument to json_encode() with array( … ).