I have a problem with my php script through which i need to send data to db. When i send data, i get blank screen, no matter what I write, select and everything else.
<?php
$con = mysql_connect("localhost","root","");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("gagq", $con);
$result = mysql_query("SELECT * FROM gagq.user WHERE id=1;");
echo $result;
mysql_close($con);
?>
basically , i get blank screen from this. i access the file through the form, submit button. Method post, action name of file. Now, i've filled the db on id=1, but no matter what i write, on php nothing works. Any idea?
try this
$result = mysql_query("SELECT * FROM gagq.user WHERE id=1;");
$array = mysql_fetch_assoc($result);
var_dump($array);
When you are in a testing phase, you should include:
/* Turns Error Reporting ON */
ini_set('display_errors',1);
error_reporting(E_ALL);
in order to see and correct any php errors, rather than adding die checking everywhere.
Related
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'
I am using the following script to check a code, so when the user enters the survey code they get the survey that is associated with that code. The part that fetches the survey is working as its supposed to, but I cant seem to get the error message to come up for some reason. If I enter a wrong code or no code all on the form this posts from, all I get is a blank page.
<?php
$con = mysql_connect("myhost","myuser","mypassword;
if (!$con) {
die('Could not connect: ' . mysql_error());
}
// Select mysql db
mysql_select_db("mydb", $con);
$questionaireID = $_POST['questionaireID'];
$result = mysql_query("SELECT * FROM itsnb_questionaire WHERE questionaireID='$questionaireID'") or die(mysql_error());
while($row = mysql_fetch_array($result)) {
if (empty($row['questionaireID'])) {
echo '<h2>Sorry I cant find a quiz with that code, please recheck your code.</h2>';
} else {
$url = $row['questionaireurl'];
header('Location: '.$url.'');
}
}
?>
It will never get there, because if the resultset is empty, it'll skip the while loop.
Try this, instead, limiting to 1 record (which is what you expect) and using an if...else instead of your while (while is only required when multiple results are expected):
$sql = "SELECT *
FROM itsnb_questionaire
WHERE questionaireID = '{$questionaireID}'
LIMIT 1";
$result = mysql_query($sql) or die(mysql_error());
if ($row = mysql_fetch_array($result)) {
$url = $row['questionaireurl'];
header('Location: '.$url.'');
} else {
echo '<h2>Sorry I cant find a quiz with that code, please recheck your code.</h2>';
}
If number of returned rows is zero than you haven't found your result therefore you can display apropriate error message
try
if (mysql_num_rows($result)<1){
//error
}
My code doesn't insert any records to mysql. What is wrong? I am really confused.
I have designed a form and I want to read data from text box and send to the database.
<?php
if(isset($_post["tfname"]))
{
$id=$_post["tfid"];
$name=$_post["tfname"];
$family=$_post["tffamily"];
$mark=$_post["tfmark"];
$tel=$_post["tftell"];
$link=mysql_connect("localhost","root","");
if (!$link)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db("university",$link);
$insert="insert into student (sid,sname,sfamily,smark,stel) values ($id,'$name','$family',$mark,$tel)";
mysql_query($insert,$link);
}
mysql_close($link);
?>
You'd better to put quotation mark for id, mark and tel after values in your query. Also as #Another Code said, you must use $_POST instead of $_post in your code. Try this and tell me the result:
<?php
if(isset($_POST["tfname"])) {
$id=$_POST["tfid"];
$name=$_POST["tfname"];
$family=$_POST["tffamily"];
$mark=$_POST["tfmark"];
$tel=$_POST["tftell"];
$link=mysql_connect("localhost","root","");
if (!$link) {
die('Could not connect: ' . mysql_error());
} else {
mysql_select_db("university",$link);
$insert="insert into student
(sid,sname,sfamily,smark,stel) values
('$id','$name','$family','$mark','$tel')";
mysql_query($insert,$link) or die (mysql_error());
mysql_close($link);
}
} else {
die('tfname did not send');
}
?>
Use mysql_query($insert,$link) or die (mysql_error()); to fetch the error message.
With the code you've provided it could almost be anything - to do some tests... have you echo'd something to confirm you are even getting the tfname in POST? Does it select the database fine? Do the fields $id, $mark, and $tel need single quotes around them as well? We need to know more about where the code is not working to provide more help but that snippet appears as though it should be running, in the interim, please use some echo's to narrow down your problem!
Try to run the generated sql query in the sql query browser. Get the query by "echo $insert" statement.
change the $insert to:
$insert="insert into student (sid,sname,sfamily,smark,stel) values ($id,'".$name."','".$family."','".$mark."','".$tel."')";
further set ini_set('display_errors',1) so that php displays the error messages as required.
Lastly, whenever doing a mysql query, try to use or die(mysql_error()) in the query so that if somethings wrong with mysql or syntax, we are aware.
$q = mysql_query($query) or die(mysql_error());
I would like to add results into array and print on screen.
Here's my code and nothing is printing...could someone help me take a look.
include('config.php');
$con = mysql_connect($host, $username, $password);
if (!$con){
die('Could not connect: ' . mysql_error());
}
mysql_select_db('members', $con) or die(mysql_error()) ;
$sql = "SELECT * FROM people where status like '%married%' ";
$result = mysql_query($sql);
while($row = mysql_fetch_array($result)){
$result_array[] = $row['id']; // <-------**here**
}
return $result_array;
echo($result_array);
mysql_close($con);
You're doing a return before you do the echo/mysql_close, so the echo/close calls never get executed. Unless you've got more code around this snippet, there's no point in having that return call, as you're not actually in a function, so the return is effectively acting as an "exit()" call, since you're already at the top of the execution stack.
Firstly, change:
mysql_select_db(mtar, $con) or die(mysql_error());
To:
mysql_select_db('mtar', $con) or die(mysql_error());
Then remove return $result_array; - you don't need it and when used outside of a function it just halts execution of the script.
Finally, change
echo($result_array);
to:
print_r($result_array);
EDIT: Some additional thoughts:
You don't need parentheses around the argument to echo - it's actually more efficient to leave them out: echo $var; is quicker than echo($var);
If you're only ever going to use the id column, then don't select the whole row: use SELECT id FROM instead of SELECT * FROM.
Are you sure you need the wildcards either side of "married"? You may well do (depends on what the possible values of status are), but you probably don't. So
$sql = "SELECT id FROM people where status = 'married' ";
May be better,
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 :)