Proper insert of date in MySql date field - php

I am posting data for event in this format via ajax:
doctor=100&date=2015-04-30&rCheck=1&rDate=2015-05-12&fromTime=21%3A35&toTime=22%3A40&status=open
I am getting it like this with php
$date = $_POST['date'];
$fromHours = $_POST['fromHours'];
$fromMinutes = $_POST['fromMinutes'];
$toHours = $_POST['toHours'];
$toMinutes = $_POST['toMinutes'];
$fromTime = $_POST['fromTime'];
$toTime = $_POST['toTime'];
$start = $date.' '.$fromTime.':00';
$end = $date.' '.$toTime.':00';
$status = $_POST['status'];
$doctor = $_POST['doctor'];
if($_POST['rCheck']==1){
$repeat = $_POST['rDate'];
}else{
$repeat = '0000-00-00';
}
When I echo any of that variables I get correct result.
I am inserting data in database like this:
$query = 'INSERT INTO events (start,end,doctor,status,repeat) VALUES ("'.$start.'","'.$end.'","'.$doctor.'","'.$status.'","'.$repeat.'")';
$result = mysqli_query($db, $query) or die (mysqli_error($db));
Now main problem is $repeat because without it everything is inserted without problem but with $repeat I get this error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'repeat) VALUES ("2015-04-30 21:35:00","2015-04-30 22:40:00","100","offen","2015-' at line 1
Insert stops after fifth '-' character. Even if $repeat is 0000-00-00
Field for that in database is date field in format 0000-00-00
I really don't know where problem is.
Thank you for helping.

repeat is a MySQL reserved word
http://dev.mysql.com/doc/refman/5.5/en/reserved-words.html
Either rename it to something else; for example "repeats", or use ticks around it
(start,end,doctor,status,`repeat`)
Look at what MySQL is telling you; it's pointing it out where it starts:
...for the right syntax to use near 'repeat
^
Sidenote:
If you're still having problems inserting data into your table, you can sanitize your input(s) using mysqli_real_escape_string() or using prepared statements as stated below.
I need to point out though, that your present code is open to SQL injection.
Use prepared statements, or PDO with prepared statements, they're much safer.

Related

How to input a date into mysql database using PHP

Ok so i already have a few insert scripts working, however when i try and use this one nothing happens. I dont get an error or anything, it simply doesnt insert into the database. I have a feeling its the date parameter, however im unsure how to make it recognizable?
<?php //SETTING SESSION VARIABLES FOR ROUND 1 TEAMS AND SCORE
session_start();
if (isset($_POST['go1'])) { // MATCHUP 1
include_once 'dbcon.php';
$_SESSION['t_team1'] = $_POST['team-1'];
$_SESSION['t_team2'] = $_POST['team-2'];
$_SESSION['s_score1'] = $_POST['score-1'];
$_SESSION['s_score2'] = $_POST['score-2'];
$team1winner = mysqli_real_escape_string($conn, $_POST['team-1']);
$team2winner = mysqli_real_escape_string($conn, $_POST['team-2']);
$date1 = mysqli_real_escape_string($conn, $_POST['date-1']);
$sql = "INSERT INTO knockout (knockout_team1, knockout_team2, knockout_date)
VALUES ('$team1winner', '$team2winner', '$date1');";
header("Location: ../tables.php?tables=winner");
}
//date html
<input type="date" name"date-1" value="date" class="date">
First of all, you are putting the SQL command in a string but you are not actually executing the command.
Second of all, if you are executing the command but you are not showing it here and your dbcon.php file is working properly, then it is more likely a date format issue.
Finally, you need to execute all of your commands especially INSERT commands in prepared statements to prevent SQL injections winch is VERY important.
Here how your code should look like :
<?php //SETTING SESSION VARIABLES FOR ROUND 1 TEAMS AND SCORE
session_start();
if (isset($_POST['go1'])) { // MATCHUP 1
include_once 'dbcon.php';
$_SESSION['t_team1'] = $_POST['team-1'];
$_SESSION['t_team2'] = $_POST['team-2'];
$_SESSION['s_score1'] = $_POST['score-1'];
$_SESSION['s_score2'] = $_POST['score-2'];
$team1winner = mysqli_real_escape_string($conn, $_POST['team-1']);
$team2winner = mysqli_real_escape_string($conn, $_POST['team-2']);
$date1 = mysqli_real_escape_string($conn, $_POST['date-1']);
$TeamsStat = $conn->prepare("INSERT INTO knockout (knockout_team1, knockout_team2, knockout_date) VALUES (?, ?, ?)");
$TeamsStat->bind_param("sss", $team1winner, $team2winner, $date1);
$TeamsStat->execute();
$TeamsStat->close();
header("Location: ../tables.php?tables=winner");
}
Where $conn is the object of your database connection.
Since prepared statements doesn't support date type and since the date is not a free input value, the knockout_date column should be a string and the variable $date1 should also be a string.
Hope that helped you.
The date format in HTML is of type dd-mm-yyyy but while inserting it into database you need to make sure its YYYY-mm-dd type. So make sure you are converting it Before inserting it
Ex :
$originalDate = $yourDateVariable;
$newDate = date("d-m-Y", strtotime($originalDate));
Or Go through manual and find alternatives functions to convert it

Update and Insert into three tables simultaneously

I have a record that needs to be updated. If the update is successful, then it should insert record into three different tables. I did it with the code below,but one of the table(tab_loan_targetsave)is not inserting.I need a third eye to looked into this, as I have had a lot of pain in fathoming where the problem lies.
Pls i need assistance.Also, I welcome better approach if possible.
<?php
if(isset($_POST["savebtn"])){
$custNo = $_POST["custid"];
$transDate = $_POST["transDate"];
$grpid = $_POST["custgrp"];
$contAmount =$_POST["amtCont"];
$amount = $_POST["amount"];
$disAmount =$_POST["disbAmt"];
$savAmount =$_POST["savAmt"];
$intAmount =$_POST["intAmt"];
$postedBy = $_SESSION["staffid"];
//$preApproved =$_POST["preAmount"];
$loanRef = $_POST["refid"];
$st = "Approved";
$appDate = date("Y-m-d H:i:s");
$appBy = $_SESSION['staffid'];
$counter = 1;
$locate = $_SESSION['location'];
$insure = $_POST["insuAmt"];
$dis = $_POST["DisAmt"];
$update = mysqli_query($connection,"UPDATE tab_loan_request SET approval_status='$st',approvalDate='$appDate',approvedBy='$appBy',loanRef='$loanRef' WHERE custid='$custNo' AND RepayStatus='1'");
if($update && mysqli_affected_rows($connection)>0){
$insertTar = mysqli_query($connection,"INSERT INTO tab_loan_targetsave(custid,grpid,transactionDate,loanRef,savingAmt,status,postedBy,location,appStatus)
VALUES('$custNo','$grpid','$transDate','$loanRef,'$savAmount','Cr','$postedBy','$locate','1')");
$insertInt = mysqli_query($connection,"INSERT INTO tab_loan_interest(custid,requestAmt,transactionDate,interestFees,postedBy,loanRef,InsuranceFees,DisasterFees)VALUES(
'$custNo','$amount','$transDate','$intAmount','$postedBy','$loanRef','$insure','$dis')");
//if($insertInt){
//}if($insertTar){
$insertSav = mysqli_query($connection,"INSERT INTO tab_loan_saving(custid,grpid,transactionDate,loanRef,loanAmount,savingAmt,status,postedBy,location,appStatus)
VALUES('$custNo','$grpid','$transDate','$loanRef','$amount','0','Cr','$postedBy','$locate','1')");
}//first if
if($insertSav){
echo "<span style='font-weight:bold;color:red;'>"." Application Approval is successful!"."</span>";
}else{
//Unable to save
echo "<span style='font-weight:bold;color:black;>"."Error! Application Approval not Successful!"."</span>";
}
}else{
$custid = "";$saving=0.00;$st="";
$transDate = "";
$grpid = "";
$amount = "";
$postedBy = "";$loanRef="";
}
?>
"#Fred: See the error generated when i used mysqli_error($connection). Could you please interprete this: ErrorMessage: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '1000.00','Cr','SPL002','Ojo','1')' at line 2 – Dave"
Seeing the error generated by the suggestion I've given you to check for errors.
You're missing a quote here '$loanRef
in your query:
VALUES('$custNo','$grpid','$transDate','$loanRef , '$savAmount'...
^ right there
I suggest to escape all of your incoming data.
I.e.:
$var = mysqli_real_escape_string($connection, $_POST['var']);
and apply that same logic to all your POST arrays.
Plus, as I stated; make sure you started the session, since there is no mention of that in your question and session_start(); wasn't included in your posted code.
The session needs to be started inside all pages using sessions.
Using a prepared statement will is better.
http://php.net/manual/en/mysqli.prepare.php
http://php.net/manual/en/pdo.prepared-statements.php
which is what you really should be using.
Additional references:
http://php.net/manual/en/mysqli.error.php
http://php.net/manual/en/function.error-reporting
Also make sure there aren't any constraints in your table(s).
Dude make sure you properly escape your variables http://php.net/manual/en/mysqli.prepare.php
i would check the Table Name! make sure it is case sesntive, also just wondering if you could do something to your database design? It seems a lot of duplicate data is going into your tables. Think about a better way to organise and store that data
I got where the error is emanting from . Just because I forgot to add a single quote to one of the values. ie missing the quote- near $loanRef. No closing string. Anyway, I was able to detect that through the error message stated parameter as adviced by Fred nad Mark. Correct
$insertTar = mysqli_query($connection,"INSERT INTO tab_loan_targetsave(custid,grpid,transactionDate,loanRef,savingAmt,status,postedBy,location,appStatus)
VALUES('$custNo','$grpid','$transDate','$loanRef','$savAmount','Cr','$postedBy','$locate','1')");
Thank you all.

Insert a record with an apostrophe mysql php

I want to insert a record with an apostrophe into a MySQL database using PHP. Following is my code:
$importer_name =mysql_escape_string ($objWorksheet->getCellByColumnAndRow(1,3)->getValue());
$exporter_name = $objWorksheet->getCellByColumnAndRow(1, 3)->getValue();
$prod_quantity_unit = $objWorksheet->getCellByColumnAndRow(1,6)->getValue();
$prod_fob_value = $objWorksheet->getCellByColumnAndRow(5,6)->getValue();
$prod_quantity = $objWorksheet->getCellByColumnAndRow(1,8)->getValue();
$prod_fob_unit= $objWorksheet->getCellByColumnAndRow(5,8)->getValue();
$prod_gross_waight= $objWorksheet->getCellByColumnAndRow(1,10)->getValue();
$prod_cif_value= $objWorksheet->getCellByColumnAndRow(5,10)->getValue();
$prod_net_weight= $objWorksheet->getCellByColumnAndRow(1,12)->getValue();
$prod_cif_unit_price= $objWorksheet->getCellByColumnAndRow(5,12)->getValue();
$prod_brand= $objWorksheet->getCellByColumnAndRow(5,14)->getValue();
$hs_code = $objWorksheet->getCellByColumnAndRow(1,17)->getValue();
$shipping_date = $objWorksheet->getCellByColumnAndRow(5,17)->getValue();
$customs = $objWorksheet->getCellByColumnAndRow(1,19)->getValue();
$transport_company = $objWorksheet->getCellByColumnAndRow(5,19)->getValue();
$country_of_origin = $objWorksheet->getCellByColumnAndRow(1,21)->getValue();
$transport_mode = $objWorksheet->getCellByColumnAndRow(5,21)->getValue();
$country_of_trade = $objWorksheet->getCellByColumnAndRow(1,23)->getValue();
$hs_code_description = $objWorksheet->getCellByColumnAndRow(1,26)->getValue();
$product_description = $objWorksheet->getCellByColumnAndRow(1,28)->getValue();
$insertquery="INSERT INTO tb_peru_data
(importer_name,exporter_name,product_quantity_unit,
product_fob_unit,product_quantity,product_fob_value,
product_gross_weight,product_cif_value,
product_net_weight,product_cif_unit_price,
product_brand,shipping_hs_code,shipping_date,
shipping_customs,shipping_transport_company,
shipping_country_of_origin,shipping_transport_mode,
shipping_country_of_trade,hs_code_description,
product_description)
VALUES
('$importer_name','$exporter_name','$prod_quantity_unit',
'$prod_fob_unit','$prod_quantity','$prod_fob_value',
'$prod_gross_waight','$prod_cif_value','$prod_net_weight',
'$prod_cif_unit_price','$prod_brand','$hs_code','$shipping_date',
'$customs','$transport_company','$country_of_origin',
'$transport_mode','$country_of_trade',
'$hs_code_description','$product_description')";
mysql_query($insertquery)or die('ErrorrPERU: '.mysql_error());
/*$del="DELETE * FROM tb_excel_file";
mysql_query($del);*/
?>
This does not work, and gives the following error:
you have an error in your SQL syntax; check the manual that corresponds
to your MySQL server version for the right syntax to use near
's','12U','6','9','54',
'34.83','55.5','31.83','6.17','','7323931000','2008/04/1' at line 3
Use mysqli_real_escape_string instead of deprecated mysql_real_escape_string
This function will force you to input mysql table / database.
This way your collation will be considered while escaping
You can use real_escape_string() in PHP. You need to escape the apostrophe (that is, tell SQL that the apostrophe is to be taken literally and not as the beginning or end of a string). To add more, I'd say that you can also use PDO, but consider using addslashes($string) and stripslashes($string).

SQL syntax error edit post

getting :
You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use
near 's Creed III', description='The plot is set in a fictional
history of real ' at line 2
when trying to edit posts on a database.
heres my display and edit php:
$result = mysql_query("SELECT * FROM gallery");
while ($row = mysql_fetch_array( $result )){
// while looping thru each record…
// output each field anyway you like
$title = $row['title'] ;
$description = $row['description'];
$year = $row['year'];
$rating = $row['rating'];
$genre = $row['genre'];
$filename = $row['filename'];
$imageid = $row['imageid'];
include '../modules/edit_display.html';
}
// STEP 2: IF Update button is pressed , THEN UPDATE DB with the changes posted
if(isset($_POST['submit'])){
$thisTitle = $_POST['title'];
$thisDescription = $_POST['description'];
$thisYear = $POST['year'];
$thisRating = $POST['rating'];
$thisGenre = $POST['genre'];
$thisNewFilename = basename($_FILES['file']['name']);
$thisOneToEdit = $_POST['imageid'];
$thisfilename = $_POST['filename'];
if ($thisNewFilename == ""){
$thisNewFilename = $thisfilename ;
} else {
uploadImage();
createThumb($thisNewFilename , 120, "../uploads/thumbs120/");
}
$sql = "UPDATE gallery SET
title='$thisTitle',
description='$thisDescription',
year='$thisYear',
rating='$thisRating',
genre='$thisGenre',
filename='$thisNewFilename'
WHERE
imageid= $thisOneToEdit";
$result = mysql_query($sql) or die (mysql_error());
}
You're suffering from an imminent dose of SQL Injection due to using a dangerous user input model.
When you type "Assassin's Creed III" in the title field, that gets placed in single quotes in the UPDATE statement in your code (via the $_POST['title'] variable):
'Assassin's Creed III'
The problem there is that MySQL sees it as 'Assassin', followed by s Creed III'. It doesn't know what to do with the latter.
Of course, this becomes a HUGE problem if someone types in valid SQL at that point, but not what you expected. Have a look at How can I prevent SQL injection in PHP? or any of several other advices on avoiding SQL Injection.
i have seen you are adding ' into database so you need to escape it using addslashes()
addslashes($thisTitle)
You have syntax error here. Use $_POST instead of $POST.
Replace
$thisYear = $POST['year'];
$thisRating = $POST['rating'];
$thisGenre = $POST['genre'];
With
$thisYear = $_POST['year'];
$thisRating = $_POST['rating'];
$thisGenre = $_POST['genre'];
you need to escape your input like
$thisDescription = mysql_real_escape_string($_POST['description']);
do this for all input that contains quotation marks etc..
NOTE: mysql will soon be gone so its advised to write new code using mysqli instead
You have alot of issues in your script.
You're trying to add ' character to database, you need to escape it properly with addslashes.
You're vulnerable to SQL Injection. Escape it properly with mysql_real_escape_string, or even better, use PDO.
Third, it is $_POST, not $POST. You're using it wrong in some areas.
Add quotes to $thisOneToEdit in query.
The error is causing because you're trying to add Assasin's Creed III string to database. The single quote breaks your query and creates a syntax error.
Do a addslashes() on the values that might contain single or double quotes like below before using them in query
$thisTitle = addslashes($_POST['title']);

Wrong mysql query in php file?

I'm trying to insert some data into my mysql database. The connection is working fine but im having a problem with sending the query correctly to the database. Below you can find the code in my php file. I also post what for type of fields they are in the Database.
Fields in the mysql database:
Reservaties_id = int
Materialen_id = int
aantal = int
effectief_gebruikt = tinyint
opmerking = Varchar2
datum_van = date
datum_tot = date
$resID = $_REQUEST['resID'];
$materialen_id = $_REQUEST['materialen_id'];
$aantal = $_REQUEST['aantal'];
$effectief_gebruikt = $_REQUEST['effectief_gebruikt'];
$opmerking = $_REQUEST['opmerking'];
$datum_van = date('YYYY-MM-DD',$_REQUEST['datum_van']);
$datum_tot = date('YYYY-MM-DD',$_REQUEST['datum_tot']);
$string = "INSERT INTO `materialen_per_reservatie`(`reservaties_id`, `materialen_id`, `aantal`, `effectief_gebruikt`, `opmerking`, `datum_van`, `datum_tot`) VALUES ($resID, $materialen_id, $aantal, $effectief_gebruikt, '$opmerking', $datum_van, $datum_tot)";
mysql_query($string);
you have to include single quotes for the date fields '$dataum_van'
$string = "INSERT INTO `materialen_per_reservatie`(reservaties_id, materialen_id, aantal, effectief_gebruikt, opmerking, datum_van, datum_tot) VALUES ($resID, $materialen_id, $aantal, $effectief_gebruikt, '$opmerking', '$datum_van', '$datum_tot')";
and this is only a example query, while implementing don't forget to sanitize your inputs
Your code has some serious problems that you should fix. For one, it is not doing any error checking, so it's no surprise the query breaks silently when it fails. Check for errors and it will tell you what goes wrong - how to do it is outlined in the manual on mysql_query() or in this reference question.. Example:
$result = mysql_query($string);
// Bail out on error
if (!$result)
{
trigger_error("Database error: ".mysql_error(), E_USER_ERROR);
die();
}
In this specific case, I'm fairly sure it's because you are not putting your values into quotes after the VALUES keyword.
Also, the code you show is vulnerable to SQL injection. You need to escape every value you use like so:
$resID = mysql_real_escape_string($_REQUEST['resID']);
for this to work, you need to put every value in your query into quotes.
try this
$string = "INSERT INTO `materialen_per_reservatie`(`reservaties_id`) VALUES ('".$resID."')";

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