Have the following code that's executed when a script is ran. (I've just changed the login for display purposes).
<?php
$conn = mysql_connect("localhost", "root", "pw123");
mysql_select_db("test_db", $conn);
$sql = "INSERT INTO test_table (fname)
VALUES ('$fname')";
mysql_query($sql);
mysql_close($conn);
?>
I've edited the code down slightly so it doesn't show every value I'm trying to enter, but essentially, everything is entering as a blank value, or in the case of numerical inputs is defaulting to 0. I can't seem to figure out why this is. The variables are definitely not blank before hand as I've got them out putting on the web page to test as such.
For reference I assign $fname a value when the input box is changed using :
fname = $("#fname").val();
(Posted on behalf of OP):
Solved this myself anyway, instead of executing the MySQL statements in the initial page that user enters data, I moved it to the secondary web page, which opens once a user has submitted their information.
$fname is empty in your script and you need declarate the variable before:
$fname = 'David';
$sql = "INSERT INTO test_table (fname) VALUES ('$fname')";
:)
Related
I am currently looking to run a basic insert query using PHP to submit HTML form data to MySQL database.
Unfortunately however the insert process isnt running.
In my Insert syntax I have tried including $_POST[fieldname], ive tried including variables as below, and ive even played around with different apostrphes but nothing seems to be working.
as a side dish, im also getting truck load of wamp deprication errors which is overwhelming, ive disabled in php.ini and php for apache.ini file and still coming up.
If anyone can advise what is wrong with my insert and anything else id be much thankful.
Ill keep this intro straightfoward.
Person logs in, if they try to get in without login they go back to login page to login.
I connect to database using external config file to save me updating in 50 places when hosting elsewhere.
Config file is working fine so not shown below.
database is called mydb.
Im storing the text field items into variables, then using the variables in the insert query.
unitID is an auto increment field so I leave that blank when running the insert.
Unfortunately nothing is going in to the mysql database.
Thanks in advance.
PS the text fieldnames are all correctly matched up
<?php
//Start the session
session_start();
//check the user is logged in
if (!(isset($_SESSION['Username']) )) {
header ("Location: LoginPage.php?i=1");
exit();
}
//Connect to the database
include 'config.php';
$UserName = $_SESSION['Username'];
$UserIdentification = $_SESSION['UserID'];
if(isset($_GET['i'])){
if($_GET['i'] == '1'){
$tblName="sightings";
//Form Values into store
$loco =$_POST['txtloco'];
$where =$_POST['txtwhere'];
$when =$_POST['txtdate'];
$time =$_POST['txttime'];
$origin =$_POST['txtorigin'];
$dest =$_POST['txtdest'];
$headcode =$_POST['txtheadcode'];
$sql= "INSERT INTO sightings (unitID, Class, Sighted, Date, Time, Origin, Destination, Headcode, UserID) VALUES ('','$loco', '$where', '$when', '$time', '$origin', '$dest', '$headcode', '$UserIdentification')";
mysql_select_db('mydb');
$result=mysql_query($sql, $db);
if($result){
$allocationsuccess = "Save Successful";
header ('Refresh: 2; url= create.php');
}
else {
$allocationsuccess = "The submission failed :(";
}
}
}
?>
"unitID is an auto increment field so I leave that blank when running
the insert"
That's not how it works. You have to omit it completely from the INSERT statement. The code thinks you're trying to set that field to a blank string, which is not allowed.
$sql= "INSERT INTO sightings (Class, Sighted, Date, Time, Origin, Destination, Headcode, UserID) VALUES ('$loco', '$where', '$when', '$time', '$origin', '$dest', '$headcode', '$UserIdentification')";
should fix that particular issue. MySQL will generate a value automatically for the field and insert it for you when it creates the row.
If your code had been logging the message produced by mysql_error() whenever mysql_query() returns false then you'd have seen an error being generated by your query, which might have given you a clue as to what was happening.
P.S. As mentioned in the comments, you need to re-write your code with a newer mysql code library and better techniques including parameterisation, to avoid the various vulnerabilities you're currently exposed to.
i am having an issue with my code...im not sure what is causing it to be entered in twice in my database. the information inputted in the form is inserted in the database and then there is another entry below it but it is empty. it happens everytime the form is submitted. how do i fix it?
<?php
// form data //
$id = $_POST['id'];
$joined = date("Y/m/d");
$uname = $_POST['uname'];
$email = $_POST['email'];
// Create connection
$connect = mysql_connect("localhost","user","pass");
mysql_select_db("db"); //select database
//register into database
$sql = mysql_query("
INSERT INTO table VALUES
('','$joined','$uname','$email')
");
echo "thanks for signing up!";
?>
I suspect this might have something to do with .htaccess redirect the second time which might explain the insert but with blank values. Please verify that the values are not null before inserting. This will keep your blank values away.
Also, check your .htaccess and fix whatever problem is causing it to hit your page twice.
There is no need to mention the id as blank as the query automatically fetches the id which is defined as primary key and auto increment.You query should like this and after the tablename you can also specify the column name from the database:
$sql=mysql_query("insert into `table`(joined,uname,column) values('$joined','$uname','$email')");
You must use validation in your form to fix duplicate date in the database.You use the id field in your form so definitely you should use of javascript.
Okay so this problem is really boggeling my mind... I have a MYSQL query I want to make so that my php program can access and update the database with lat and long coordinates of a user and im getting issues...
This is non working code:
$currUsername = strtolower($_SESSION['username']);
$sql= "UPDATE users SET pos_Lat=$latitude, pos_Long=$longitude WHERE username=$currUsername";
$result = mysql_query($sql, $link);
The working code
$currUsername = "email_that_is_returned"
$sql= "UPDATE users SET pos_Lat=$latitude, pos_Long=$longitude WHERE username=$currUsername";
$result = mysql_query($sql, $link);
Is this because session returns data that is not able to be placed inside a query?
Check whether the session was started or not. if not started then add the following code to your page and then check its working or not.. i thing your session does not return any value.. so start session by using the code session_start();
session_start();
$currUsername = strtolower($_SESSION['username']);
$sql= "UPDATE users SET pos_Lat=$latitude, pos_Long=$longitude WHERE username=$currUsername";
$result = mysql_query($sql, $link);
You can check what type of data it is returning.
print $_SESSION['username'].
Also there is a chance to break the SQL query if the $_SESSION['username'] returns data with spaces. Make sure the SQL query not failing even if the $_SESSION['username'] contains spaces and singlequotes etc..
Kinda new to mysql and php
I have a hit counter for each page on my site and a private page that list all pages and hits.
I have a button that will reset all pages to zero and next to each page listing I have a reset button that will reset each page individually. This all was using a text file but now I am swtching to mysql database. I have coded the "RESET ALL" button to work but can not get the individual page buttons to work.
the processing code is:
if($_POST[ind_reset]) {
$ind_reset = $_POST[ind_reset];
mysql_connect("server", "username", "password") or die(mysql_error());
mysql_select_db("database") or die(mysql_error());
$sql = 'UPDATE counters SET Hits =\'0\' WHERE Page = \'$ind_reset\';';
}
and the html form code is a string:
$page_reset = "<form id='Reset' action='counter_update.php' method='post'>
<button type='submit' name='ind_reset' value='$formPage'>RESET</button>
</form>";
Let's start with the first thing:
if($_POST[ind_reset]) {
should be
if($_POST['ind_reset']) {
It works without quotes because PHP is silently correcting your error. If you turned error reporting to E_ALL, you would get to see the error message.
One thing that you need to consider is that you can never trust POST data to be what you think it's supposed to be. Maybe you put in a typo. Maybe a hacker is sending you fake POST data. Whichever it is, it will mess up your code if the wrong thing gets put in that database update. For this reason, instead of simply plugging in that POST value into your database, you should have a checker to make sure that the value is a valid one. When I do things like this, I make an array of possible values and use only those values when updating or inserting into the database. Example:
$pages = array('value_on_page'=>'value_put_in_database',
'xyz'=>'thing_in_database_2');
//the valid things to post are either 'value_on_page' or 'xyz',
//but what goes into the database are the values those keys point to
//e.g. if $_POST['ind_reset'] == 'xyz', $ind_reset will be 'thing_in_database_2'
$key = $_POST['ind_reset'];
if(!isset($pages[$key])) {
//if that posted value isn't a key in the array, it's bad
error_log('Invalid posted page'.$key);
} else {
//this is a valid posted page
$ind_reset = $pages[$key];
//** do the database stuff right here in this spot **//
}
Now, for the reason your posted code doesn't work, you are missing the final, crucial part of doing a database query: the part where you actually run the query.
$conn = mysql_connect("server", "username", "password") or error_log(mysql_error());
mysql_select_db("database") or error_log(mysql_error());
$sql = 'UPDATE counters SET Hits =\'0\' WHERE Page = \'$ind_reset\';';
mysql_query($sql, $conn) or error_log(mysql_error());
I hope you have noted that I replaced "die" with "error_log." If you do error_log(mysql_error(), 1, 'youremail#example.com'), it will email it to you. Otherwise, as with in my examples, it gets put into wherever your system's error log file is. You can then have a nice history of your database errors so that, when you inevitably return to StackOverflow with more questions, you can tell us exactly what's been going on. If you use a file, just make sure to either rotate the error log file's name (I name them according to the day's date) or clear it out regularly, or it can get really, really long.
Using the mysqli code you posted in your comment is a better idea than the mysql_* functions, but you don't quite have it correct. The "bind_param" part sticks your variable into the spot where the question mark is. If your variable is a string, you put "s" first, or if it's an integer, you put "i" first, etc. And make sure you close things once you're done with them.
$db = new mysqli("server", "username", "password", "database");
if(!$db->connect_errno) {
$stmt = $db->prepare("UPDATE counters SET Hits = '0' where Page = ?");
$stmt->bind_param('s',$ind_reset); //assuming $ind_reset is a string
if(!$stmt->execute()) {
error_log($stmt->error);
}
$stmt->close();
} else {
error_log($db->connect_error);
}
$db->close();
This is driving me nuts. I am using the jQuery image upload and crop from
http://www.webmotionuk.co.uk/php-jquery-image-upload-and-crop/
I am using a modified version of the suggestion on here to store the file location in a MySQL database. The mod is that I use INSERT on a table it works great except one thing, the 'owner' variable $id is being stored as $id and not as the value of $id. I can echo the value if $id on each $_POST so I know it's there.
I am pretty sure my syntax is correct but I don't understand why it is doing this.
$cropped = resizeThumbnailImage($thumb_image_location, $large_image_location,$w,$h,$x1,$y1,$scale);
//connect to the database
include 'config.php';
// check connection
if (mysqli_connect_errno()) {
exit('Connect failed: '. mysqli_connect_error());
}
$sql = "INSERT INTO `photos` (`id`,`owner`,`url`) VALUES ('id','".$id."','".$thumb_image_location."')";
// Performs the $sql query on the server to insert the values
if ($conn->query($sql) === TRUE) {
$conn->close();}
//Reload the page again to view the thumbnail
header("location:".$_SERVER["PHP_SELF"]);
exit();
The first line is 246 and the last 3 are the orginal 247-250.
Thanks for any help you can provide.
Ok, I don't know if this is my brain fart or an issue with PHP or a bit of both. I have $id assigned from the _SESSION variable in the header of each page AND (having forgotten that) I was passing $id as _POST data (same value). Once I cut out the _POST data passing and just pulled the _SESSION variable it works fine. But assigning a variable multiple times shouldn't be an issue, should it?
the query line needs to be like this:
$sql = "INSERT INTO `photos` (`id`,`owner`,`url`) VALUES ('id','$id','$thumb_image_location')";
your syntax works fine too, as seen here
this is how my syntax works, here
Note: both work the same, so still trying to figure out what's wrong in OP's code.