Im using phpgraph lib to create graphs on my linux server. I tried an example and it worked, but I had provided it with the data.
then I wanted to connect it to mysql database and plot a query, when I run it, nothing happens, I don't see any output on the page or any errors, I don't see any output on the page at all, even if I put wrong credentials to my database e.t.c any inputs?
I have executed the sql statement on sql server and it's working fine.
the version of php the server has is PHP 5.3.3
<?php
include('phpgraphlib.php');
$graph= new PHPGraphLib(550,350);
$link = mysql_connect('localhost', 'user', 'password')
or die('Could not connect: ' . mysql_error());
mysql_select_db('databasename' or die('Could not select database');
$dataArray=array();
//get data from database
$sql="my sql statement";
$result = mysql_query($sql) or die('Query failed: ' . mysql_error());
if ($result) {
while ($row = mysql_fetch_assoc($result)) {
$salesgroup=$row["var1"];
$count=$row["count"];
//add to data areray
$dataArray[$salesgroup]=$count;
}
}
//configure graph
$graph->addData($dataArray);
$graph->setTitle("Sales by Group");
$graph->setGradient("lime", "green");
$graph->setBarOutlineColor("black");
$graph->createGraph();
?>
I fixed it, I was expecting to see errors on the webpage, but didn't see any on CHROME, I hen opened it in IE and saw error 500.
Troubleshooted through the log file.
Turned out the sql statement wasn't suppose to have double quotes e.g instead of
where name="john"
it's suppose to be
where name='john'
Related
I was uploading data from my android application to a PHP file then inserting it to Mysql database, the problem i am having that i buy a new hosting plan and when i configured everything in the new hosting, i try to upload data from the application it shows only blank fields in the table, i am sure that there's no problem in the PHP or the android code, cause it was working fine and great with the old hosting.. i tried to change the encoding but same issue.
Here's the PHP file:
<?php
$con = mysql_connect("HOST","USER","PASS");
if (!$con)
{
die('Could not Connect:'. mysql_error());
}
mysql_select_db("TABLE",$con);
mysql_query ("INSERT INTO table (rep_desc,dateT) VALUES ('".$_REQUEST['report_Desc']."','".$_REQUEST['Date_Time']."')");
mysql_close($con);
?>
Thanks in advance
If there is any auto_increment column in table. Better check if its auto_increment flag not got uncheck.
Please also check length of database fields and data that you actually trying to insert.
Name of all request variables, columns, tables should be in proper case if it is a UNIX system.
get all the tables by something like this query ...
$sql = "SHOW TABLES FROM $dbname";
$result = mysql_query($sql);
if (!$result) {
echo "DB Error, could not list tables\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
while ($row = mysql_fetch_row($result)) {
echo "Table: {$row[0]}\n";
}
Here is the code:
<?php
$con=mysql_connect('localhost', 'itorras', 'passwordhere');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db( "my_db" ) or die( 'Error'. mysql_error() );
$sql="INSERT INTO Brackets(have things here)
VALUES(and here)";
if (!mysqli_query($con,$sql))
{
die('Error with adding row: ' . mysqli_error($con));
}
echo "Thank you, your bracket has been submited.";
echo "<a href='index.html'>Click here to go back to home page</a>";
?>
I am using the username and password that is given the rights to mess with the database.
So when i try to submit something into the file none of the top errors run but then the bottom error prints error with adding row and nothing more. This file worked fine on my local server but has not worked on the web host. I am using godaddyweb hosting, so phpmyadmin and cpanelx.
If you need any more info let me know. Have been at this for a couple hours.
You are using mysql_connect but mysqli_query. You are mixing mysql and mysqli. Use only mysqli_*.
It appears you are using two different libraires at the same time
The original mysql API is deprecated for various reasons and you should not use it in new code
Instead, use PDO or mysqli
In your code, you are using the original mysql for the db connect, and then you use mysqli for the query. Instead you should only use mysqli
Replace the following code :
$con=mysql_connect('localhost', 'itorras', 'passwordhere');
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
With :
$db = new mysqli('localhost', 'user', 'pass', 'demo');
if($db->connect_errno > 0){
die('Unable to connect to database [' . $db->connect_error . ']');
}
Then you can query the database with something similar :
$sql = <<<SQL
SELECT *
FROM `users`
WHERE `live` = 1
SQL;
if(!$result = $db->query($sql)){
die('There was an error running the query [' . $db->error . ']');
}
For a complete overview, you can take a look at the documentation
http://ca2.php.net/mysqli
I wonder whether someone can help me please.
I'm trying to put together a PHP script that takes data from an xml file and places the data in a mySQL data. I've been working on this for a few days and I'm still can't seem to get this right.
This is the code that I've managed to put together:
<?
$objDOM = new DOMDocument();
$objDOM->load("xmlfile.xml");
$Details = $objDOM->getElementsByTagName("Details");
foreach( $Details as $value )
{
$listentry = $value->getElementsByTagName("listentry");
$listentrys = $listentry->item(0)->nodeValue;
$sitetype = $value->getElementsByTagName("sitetype");
$sitetypes = $sitetype->item(0)->nodeValue;
$sitedescription = $value->getElementsByTagName("sitedescription");
$sitedescriptions = $sitedescription->item(0)->nodeValue;
$siteosgb36lat = $value->getElementsByTagName("siteosgb36lat");
$siteosgb36lats = $siteosgb36lat->item(0)->nodeValue;
$siteosgb36lon = $value->getElementsByTagName("siteosgb36lon");
$siteosgb36lons = $siteosgb36lon->item(0)->nodeValue;
//echo "$listentrys :: $sitetypes :: $sitedescriptions :: $siteosgb36lats :: $siteosgb36lons <br>";
}
require("phpfile.php");
//Opens a connection to a MySQL server
$connection = mysql_connect ("hostname", $username, $password);
if (!$connection) {
die('Not connected : ' . mysql_error());
}
// Set the active MySQL database
$db_selected = mysql_select_db($database, $connection);
if (!$db_selected) {
die ('Can\'t use db : ' . mysql_error());
}
mysql_query("INSERT IGNORE INTO scheduledsites (listentry, sitetype, sitedescription, siteosgb36lat, siteosgb36lon) VALUES('$listentrys','$sitetypes','$sitedescriptions','$siteosgb36lats','$siteosgb36lons') ")
or die(mysql_error());
echo "Data Inserted!";
?>
I can pull the data from the xml file, but it's the part of the script that sends the data to my database table that I'm having trouble with.
The script runs but only the last record is saved to the database.
I can parse the fields from the xml file without any problems and the check I'm trying to put in place is, if there is a 'listentry' number in the new data that is matched to one already in the table then I don't want that record to be added to the table, i.e. ignore it.
I just wondered whether someone could perhaps take a look at this please and let me know where I'm going wrong.
Many thanks
You are only calling mysql_query once. So it will only insert one row.
The sql needs to be inside the loop.
So a friend of mine and I are using both xampp on ubuntu, if that helps, to connect between each other's website, We both created the same php file to connect, so we use de IP of the other, but then it says an error
Warning: mysql_connect() [function.mysql-connect]: Host 'coke-laptop.local' is not allowed to connect to this MySQL server in /opt/lampp/htdocs/connection.php on line 2
Could not connect: Host 'coke-laptop.local' is not allowed to connect to this MySQL server
We have this code on the connection.php file:
<?php
$link = mysql_connect('10.100.161.37','root','');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
//echo 'Connected successfully';
$db_selected = mysql_select_db('Prueba', $link);
if (!$db_selected) {
die ('Can\'t use Prueba : ' . mysql_error());
}
// This could be supplied by a user, for example
$firstname = 'fred';
$lastname = 'fox';
// Formulate Query
// This is the best way to perform an SQL query
// For more examples, see mysql_real_escape_string()
$query = sprintf("SELECT * FROM Agencia");
// Perform Query
$result = mysql_query($query);
// Check result
// This shows the actual query sent to MySQL, and the error. Useful for debugging.
if (!$result) {
$message = 'Invalid query: ' . mysql_error() . "\n";
$message .= 'Whole query: ' . $query;
die($message);
}
// Use result
// Attempting to print $result won't allow access to information in the resource
// One of the mysql result functions must be used
// See also mysql_result(), mysql_fetch_array(), mysql_fetch_row(), etc.
while ($row = mysql_fetch_assoc($result)) {
echo $row['ID'] . " ";
echo $row['Nombre'] . "\n\r";
}
// Free the resources associated with the result set
// This is done automatically at the end of the script
mysql_free_result($result);
mysql_close($link);
?>
If we use the IP just like that, we can enter each others xampp normal welcome page.
Check you have enabled remote access to the MySQL server. Open the my.cnf file (probably found inside xampp/etc/), go to the [mysqld] section and add the following (using your own ip address instead of the example)
bind-address=192.168.1.100
If there is a line that says skip-networking, comment that out so it looks like this:
# skip-networking
then restart the MySQL server
It looks like your MySQL database isn't allowing you to connect remotely with the credentials you provided. You will need to configure a remote user to connect. Try looking into MySQL Grants.
For Example:
GRANT SELECT, INSERT ON database.* TO 'someuser'#'somehost';
I'm just starting out writing this code and when I add in the db select and refresh the page, instead of showing all the other html on the page or an error, it just shows up blank.
Here's what I've got-
$link = mysql_connect('vps2.foo.com:3306', 'remote_vhost30', 'password');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
$db_selected = mysql_select_db('best_of', $link);
if (!$db_selected) {
die ('Can\'t use foo : ' . mysql_error());
}
$sections_result = "SELECT * FROM sections";
$sections_query = mysql_query($sections_result) or die(mysql_error());
$sections_array = mysql_fetch_array($sections_result) or die(mysql_error());
This code above returns a blank page. If I comment out the row starting with $db_selected the page loads fine. Obviously it doesn't do anything with the data but no errors.
What's the problem? (And yes, I am connecting to a remote server, but the $link produces no errors)
The last line of code should be:
$sections_array = mysql_fetch_array($sections_query) or die(mysql_error());
You are trying to fetch rows from the variable $sections_result, which is your query string and not the result set.
Turn on error reporting, with error_reporting(E_ALL) like mentioned in one of the other answers.
Incidentally, I suspect the problem is that PHP is throwing an error, but you've disabled the display of errors - hence the display of a blank white page. Check the status of 'display_errors' within your php.ini file.
NB: If this is a production server, you should leave display_errors set to off.
Check it really is that line by replacing this:
$db_selected = mysql_select_db('best_of', $link);
With this:
if (! $db_selected = mysql_select_db('best_of', $link)) die('Unable to select database');
As MitMaro says you've muddled _result and _query. This might be better:
$sections_query = "SELECT * FROM sections";
$sections_result = mysql_query($sections_query) or die(mysql_error());
$sections_array = mysql_fetch_array($sections_result) or die(mysql_error());
Hope that helps :)