I am trying to insert values for longitude, latitude and descriptions of locations generated from google maps (the ID value is to be generated within PhpMyAdmin. When I run the following code I always get the "error" message. If I "echo $sql;" before //insert values into database then I get the intended output so I guess it must be a problem with transferring to the database.
I'm using phpMyAdmin from within MAMP so am not sure if that is causing any issues of whether I'm missing something obvious in the code? I'm fairly new to PHP so may have missed something obvious! Any help would be much appreciated.
<?php
$con=mysqli_connect("localhost", "root", "root", "googlemaps") or die("could not connent to db");
error_reporting(0);
for($i=0;$i<count($_POST['value']);$i=$i+3)
{
$sql .= "(NULL, '".$_POST['value'][$i]."', '".$_POST['value'][$i+1]."', '".$_POST['value'][$i+2]."'),";
}
//remove last comma
$sql = substr($sql,0,-1);
//insert values into database
$query = "INSERT INTO 'googlemaps'.'values' ('id', 'lat', 'lng', 'des') VALUES " .$sql;
$status = mysqli_query($con, $query);
if($status)
echo "inserted";
else
echo "error";
?>
Related
Once again I come back to all of you with another question.
I have tried everything in my mind as well as most of the recommendations I have found on the web and here in Stackoverflow but nothing seems to fix this issue for me.
For some reason the sql command in my code is returning false even though it should not.
Here is my php file called (dbRKS-DBTest.php)
<?php
//Gets server connection credentials stored in serConCred.php
//require_once('/../prctrc/servConCred2.php');
require_once('C:\wamp64.2\www\servConCred2.php');
//SQL code for connection w/ error control
$con = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);
if(!$con){
die('Could not connect: ' . mysqli_connect_error());
}
//Selection of the databse w/ error control
$db_selected = mysqli_select_db($con, DB_NAME);
if(!$db_selected){
die('Can not use ' . DB_NAME . ': ' . mysqli_error($con));
}
//VARIABLES & CONSTANTS
//Principal Investigator Information
$PI_Selected = '6';
//Regulatory Knowledge and Support Core Requests variables
$RKS_REQ_1_Develop = '1';
//This sets a starting point in the rollback process in case of errors along the code
$success = true; //Flag to determine success of transaction
//start transaction
$command = "SET AUTOCOMMIT = 0";
$result = mysqli_query($con, $command);
$command = "BEGIN";
$result = mysqli_query($con, $command);
//Delete this portion of code afyer testing is finished
//Core Requests saved to database
$sql = "INSERT INTO rpgp_form_table_3 (idPI, RKS_REQ_1_Develop)
VALUES ('$PI_Selected', '$RKS_REQ_1_Develop')";
//*************TEsts code for "SCOPE_IDENTITY()" -> insert_id() for mysql
$sqlInsertId = mysqli_insert_id($con); //This value is supposed to be 0 since no queries have been executed.
echo "<br>MYSQLi_INSERT_ID() value before query should be 0 and it is:= " . $sqlInsertId;
//Checks for errors in the db connection.
$result = mysqli_query($con, $sql); //Executes query.
if($result == false){ //Checks to see for errors in previews query ($sql)
//die ('<br>Error in query to Main Form: Research Proposal Grant Preparation: ' . mysqli_error($con));
echo "<br>Result for the sql run returned FALSE. Check for error in sql code execution.";
echo "<br>Error given by php is: " . mysqli_error($con);
$success = false; //Chances success to false is it encounted an error in order to rollback transaction to database
}
else{
//*************TEsts code for "SCOPE_IDENTITY()" -> insert_id() for mysql
$sqlInsertId = mysqli_insert_id($con); //Saves the last id entered. This would be for the main table
echo "<br>MYSQLi_INSERT_ID() value after Main form query= " . $sqlInsertId; //Displays id last stored. This is the main forms id
$MAIN_ID = mysqli_insert_id($con); //Sets last entered id in the MAIN Form db to variable
}
//Checks for errors or craches inside the code
// If found, execute rollback
if($success){
$command = "COMMIT";
$result = mysqli_query($con, $command);
echo "<br>Tables have been saved witn 0 errors.";
}
else{
$command = "ROLLBACK";
$result = mysqli_query($con, $command);
echo "<br>Error! Databases could not be saved. <br>
We apologize for any inconvenience this may cause. <br>
Please contact a system administrator at PRCTRC.";
}
$command = "SET AUTOCOMMIT = 1"; //return to autocommit
$result = mysqli_query($con, $command);
//Displays message
//echo '<br>Connection Successfully. ';
//echo '<br>Database have been saved';
//Close the sql connection to dababase
mysqli_close($con);
?>
Here is my php frontend html code named (RPGPHomeQueryTest.php)
<!DOCTYPE html>
<html>
<head>
<meta charset="UTF-8">
</head>
<form id="testQuery" name="testQuery" method="post" action="../dbRKS-DBTest.php" enctype = "multipart/form-data">
<input type="submit" value="Submit query"/>
</form>
</html>
And here is how my database looks (rpgp_form_table_3):
So, when I open my html code, All I will see is a button since its all the code there is there. Once you press the button, the form should submit and execute the php code called (dbRKS-DBTest.php). This should take the predetermine values I already declared and saved them to the database called (rpgp_form_table_3). This database is set to InnoDB format.
Now, the output I should be getting is a message saying "Tables have been saved witn 0 errors." but the problem is that the message I am getting is this one bolow:
I honestly don't know why. I am posting this message to find guidance to this issue. I am still learning by myself and its been very did-heartedly to not find a solution this fixing this.
As always, I thank you for your patient and guidance! Let me know what other details I can provide.
Here is the SQL code you run:
$sql = "INSERT INTO rpgp_form_table_3 (idPI, RKS_REQ_1_Develop)
VALUES ('$PI_Selected', '$RKS_REQ_1_Develop')";
You are inserting data into rpgp_form_table_3. From the screenshot, we can see that table has several (7) fields yet you are only inserting 2 fields. The question then is: do you need to specify a value for all fields?
The error you are getting states
Error given by php is: Field 'idCollaRecord_1' doesn't have a default value Error! Databases could not be saved.
It's clear that you have to insert the row by specifying a value for each column, not just the two columns you are interested in.
Try
$sql = "INSERT INTO rpgp_form_table_3 (idPl, RKS_REQ_1_Develop, idCollaRecord_1, idCollaRecord_2, idCollaRecord_3, idCollaRecord_4)
VALUES ('$PI_Selected', '$RKS_REQ_1_Develop',0,0,0,0)";
Try this insert code. If the PI_Selected is NUMERIC use the First one. If it is string use the second one
$sql = "INSERT INTO rpgp_form_table_3 (idPI, RKS_REQ_1_Develop) VALUES (" .
$PI_Selected . ",'" . $RKS_REQ_1_Develop . "')";
$sql = "INSERT INTO rpgp_form_table_3 (idPI, RKS_REQ_1_Develop) VALUES ('" .
$PI_Selected . "','" . $RKS_REQ_1_Develop . "')";
I've updated my project to it's almost final form and made a small change. The suggestion of #Raghbendra Nayak worked before the change and now, I'm trying to incorporate it to my new form. After 'submit' it pushes the item to the folder still and the link to the DB, however the link is on a separate row and not on the same row where the data name-description written is found.
Update: Got it working guys. Thanks!
You can simply follow the below code to insert data in database why you are using mysqli_connect to run the insert query.
Updated:
Change your if condition:
if(move_uploaded_file($file['tmp_name'],$upload_directory.$path)){
mysqli_connect("localhost", "root", "", "order") or die ("Connecting to DB failed");
mysqli_connect("INSERT INTO item VALUES ('', '$path')");
}
To
if(move_uploaded_file($_FILES['filename']['tmp_name'],$upload_directory.$path)){
$conn = mysqli_connect("localhost", "root", "", "order") or die ("Connecting to DB failed");
// If you want to save image name you can get like below:
$filename = $_FILES["filename"]["name"];
$path = $path."/".$filename;
$sql = "INSERT INTO item (path) VALUES ('$path')";
if ($conn->query($sql) === TRUE) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . $conn->error;
}
}
You have to add your image name also. Hope it will work for you, let me know it anything required.
I have a page which allows a user to create an entry and insert it into the database via submitting a form through PHP.
I have had the page working in the past, and it worked perfectly, but now for some reason it has stopped working. Basically, I submit the form, and there are no error messages, but my entry does not appear in the database.
Here is my PHP:
<?php // add_entry.php
session_start();
include_once 'creds.php';
$con=mysqli_connect("$db_hostname","$db_username","$db_password","$db_database");
if (isset($_POST['group'])){
$lab = $_SESSION['labname'];
$author = $_SESSION['username'];
$name = $_POST['entryName'];
$group = $_POST['group'];
$protocol = $_POST['protocol'];
$permission = $_POST['permission'];
$array = $_POST['protocolValues'];
// $filearray = $_POST['fileArray']; Don't forget to add '$filearray', to your query after ARRAY.
$project = $_POST['project'];
$query = "INSERT INTO data (protocol, name, lab, author, uniquearray, usergroup, project, permissionflag) VALUES ('$protocol', '$name', '$lab', '$author', '$array', '$group', 'project', '$permission')";
mysqli_query($con, $query);
mysqli_close($con);
}else{
echo "FAILURE. GROUP WAS NOT SET.";
}
?>
Any ideas as to why this may be happening?
Thanks in advance!
In addition to the accepted answer which I applaud, this 'project' in your VALUES()
is missing the $ sign - therefore, it's not being passed as a variable but as a simple string.
so change it to '$project'
otherwise, the value entered in DB will be project and not the intended POST value from your form.
You're also not checking for errors, which is crucial during the development stages of a project.
Add error reporting to the top of your file(s)
error_reporting(E_ALL);
ini_set('display_errors', 1);
which will signal any errors found in your code.
try to add
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
after $con=mysqli_connect(...);
and replace this
mysqli_query($con, $query);
with this
if (!mysqli_query($con, $query)) {
printf("Errormessage: %s\n", mysqli_error($con));
}
you will be able to debug the problem (if related to the query)
I am not sure what I am doing wrong, can anybody tell me?
I have one variable - $tally5 - that I want to insert into database jdixon_WC14 table called PREDICTIONS - the field is called TOTAL_POINTS (int 11 with 0 as the default)
Here is the code I am using. I have made sure that the variable $tally5 is being calculated correctly, but the database won't update. I got the following from an online tutorial after trying one that used mysqli, but that left me a scary error I didn't understand at all :)
if(! get_magic_quotes_gpc() )
{
$points = addslashes ($tally5);
}
else
{
$points = $tally5;
}
$sql = "INSERT INTO PREDICTIONS ".
"(TOTAL_POINTS) ".
"VALUES('$points', NOW())";
mysql_select_db('jdixon_WC14');
I amended it to suit my variable name, but I am sure I have really botched this up!
help! :)
I think you just need to learn more about PHP and its relation with MYSQL. I will share a simple example of insertion into a mysql database.
<?php
$con=mysqli_connect("localhost","peter","abc123","my_db");
// Check for errors in connection to database.
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$query = "INSERT INTO Persons (FirstName, LastName, Age) VALUES ('Peter', 'Griffin',35)";
mysqli_query($con, $query);
mysqli_close($con); //Close connection
?>
First, you need to connect to the database with the mysqli_connect function. Then you can do the query and close the connection
Briefly,
For every PHP function you use, look it up here first.
(You will learn that it is better to go with mysqli).
http://www.php.net/manual/en/ <---use the search feature
Try working on the SQL statement first. If you have the INSERT process down, proceed.
You need to use mysql_connect() before using mysql_select_db()
Once you have a connection and have selected a database, now you my run a query
with mysql_query()
When you get more advanced, you'll learn how to integrate error checking and response into the connection, database selection, and query routines. Convert to mysqli or other solutions that are not going to be deprecated soon (it is all in the PHP manual). Good luck!
if(! get_magic_quotes_gpc() )
{
$points = addslashes ($tally5);
}
else
{
$points = $tally5;
}
mysql_select_db('jdixon_WC14');
$sql = "INSERT INTO PREDICTIONS (TOTAL_POINTS,DATE) ". //write your date field name instead "DATE"
"VALUES('$points', NOW())";
mysql_query($sql);
I use this code , to log $_SERVER['REMOTE_ADDR']; to my small db
my issue is value never saved to db , cant figure what i missed in the code
Any tips ?
<?php
mysql_connect("localhost", "usr", "passwd");
mysql_select_db("db") or die ( 'Can not select database' );
function initCounter() {
$ip = $_SERVER['REMOTE_ADDR'];
$sql = "INSERT INTO logs(REMOTE_ADDR,) VALUES ('$ip')";
}
echo $_SERVER['REMOTE_ADDR'];
?>
This should work. In addition to the other comments here, you had a comma (,) too much in your query.
<?php
mysql_connect("localhost", "usr", "passwd");
mysql_select_db("db") or die ( 'Can not select database' );
function initCounter() {
$ip = $_SERVER['REMOTE_ADDR'];
$sql = "INSERT INTO logs (REMOTE_ADDR) VALUES ('$ip')";
mysql_query($sql);
}
initCounter();
?>
You aren't actually executing the query. You create the SQL but don't use mysql_query($sql)
You have a comma at this point in the SQL REMOTE_ADDR, <-- remove that
When you execute the query, use mysql_error() to test for an error message (and check the result of mysql_query() for a boolean false.
Finally I would suggest switching to MySQLi or PDO.
If that's you're full code... there is one thing missing you actually need to EXECUTE the query...
mysql_query($sql);
EDIT:
I have just noticed, you're connecting to the DB OUTSIDE of the function trying to run the Query... obviously it will fail as inside the function, it has no awareness of the DB connection.