Database error when inserting values into MySql - php

I have an error on insert value mysql.
Please see my PHP code
<?php
$ali = $_POST['ali'];
$con = #mysqli_connect('localhost', 'root', '', 'mohammad');
if (!$con) {
echo "Error: " . mysqli_connect_error();
exit();
}
$insertinto_ic_add = "INSERT INTO sq (text) VALUES ('" . $ali . "')";
mysqli_query($con, $insertinto_ic_add) or die("database error:" . mysqli_error($con));
?>
<form action="" method="post">
<input name="ali">
</form>
I input the values " n't " and an error occurs:
database error:You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 't')' at line 2

I agree that this is not showing SQL injection. But the prevention for such is the same as the fix for your problem. You must escape certain characters (in particular the apostrophe) in the text.
Notice that the error message even points to the apostrophe.
If you echoed the statement, you would see
INSERT INTO sq (text)
VALUES ('blah blah don't do this')
Observe the three apostrophes, and think how confused the parser will be.
Better code would be something like
$mali = $con->real_escape_string($ali);
$insertinto_ic_add = "INSERT INTO sq (text)
VALUES ('" . $mali . "')";

Related

how to combine query for multiple able to save data

can it be combine into 1 query?
this is the query that im trying to combine? or is there a better way to relate these to table?
$insert_row = $mysqli->query("INSERT INTO orderlist
(TransactionID,ItemName,ItemNumber, ItemAmount,ItemQTY)
VALUES ('$transactionID','$itemname','$itemnumber', $ItemTotalPrice,'$itemqty')");
$insert_row1 = $mysqli->query("INSERT INTO order
(BuyerName,BuyerEmail,TransactionID)
VALUES ('$buyerName','$buyerEmail','$transactionID')");
when i run these both only one query is functional, so what im trying to do is to make them both works.
im open to any suggestion
The reason why your second query isn't working is because of the use of order and not escaping it; it is a MySQL reserved word:
https://dev.mysql.com/doc/refman/5.5/en/keywords.html
Sidenote: ORDER is used when performing a SELECT... ORDER BY...
https://dev.mysql.com/doc/refman/5.0/en/select.html
Checking for errors would have shown you the syntax error such as:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax near 'order
http://php.net/manual/en/mysqli.error.php
Therefore, wrap it in ticks:
$insert_row1 = $mysqli->query("INSERT INTO `order` ...
or rename your table to something other than a reserved word, say orders for example.
If you wish to combine both queries, you can use multi_query()
http://php.net/manual/en/mysqli.quickstart.multiple-statement.php
Example from the manual:
<?php
$mysqli = new mysqli("example.com", "user", "password", "database");
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
if (!$mysqli->query("DROP TABLE IF EXISTS test") || !$mysqli->query("CREATE TABLE test(id INT)")) {
echo "Table creation failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
$sql = "SELECT COUNT(*) AS _num FROM test; ";
$sql.= "INSERT INTO test(id) VALUES (1); ";
$sql.= "SELECT COUNT(*) AS _num FROM test; ";
if (!$mysqli->multi_query($sql)) {
echo "Multi query failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
do {
if ($res = $mysqli->store_result()) {
var_dump($res->fetch_all(MYSQLI_ASSOC));
$res->free();
}
} while ($mysqli->more_results() && $mysqli->next_result());
?>
I also need to point out that your present code may be open to SQL injection since I do not know if you are escaping your data.
If not, then use prepared statements, or PDO with prepared statements, they're much safer.
try to add IF statement.
if ($insert_row = $mysqli->query("INSERT INTO orderlist(TransactionID,ItemName,ItemNumber, ItemAmount,ItemQTY)VALUES ('$transactionID','$itemname','$itemnumber', $ItemTotalPrice,'$itemqty')"));
{
$insert_row1 = $mysqli->query("INSERT INTO order (BuyerName,BuyerEmail,TransactionID) VALUES ('$buyerName','$buyerEmail','$transactionID')");
}

Error inserting values to a table in MySql

I am having error inserting values to a database table in mysql.The connection is allright. I have checked it. My code is :
$emails = implode(",", $not_submitted);
$sql_update_query = "INSERT INTO reminders_table(id,group_name,runtimes,emails) VALUES(NULL, '".mysql_real_escape_string($group_name) ."' ,'".mysql_real_escape_string($runtimes) ."' , '".mysql_real_escape_string($emails) ."')";
mysql_query(sql_update_query, $con);
echo $sql_update_query, "<br>";
echo mysql_error(), "<br>";
After seeing the error in my console, it says :
"responseText: "INSERT INTO reminders_table(id,group_name,runtimes,emails) VALUES(NULL, 'BIT' , 'tue,wed-02:45,23:15' , 'c_faw,)<br>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 'sql_update_query' at line 1<br>"Reminders have been sent....! Please close this page."↵"
Any help is appreciated. So far I have tried debugging a lot. I added "mysql_real_escape_string" also, but still it doesn't work.
It a missing a Single quote after email variable.

Cannot find error in my php script. Can someone point it out to me

I have written a php script on my web server to insert values into the table table3. The variables used to get values are username and image. username contains varchar type data and image contains text type data in it. I need to insert it into my table table3 . The table3 is having two columns username and imagename which is of varchar type and text type respectively.
When I try to run the above script by entering values, an error shows as given below:
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 '1' at line 1.
I don't understand what the error is and I'm stuck here with knowing the error. Can someone please clear the errors for me. I'm a newbie in php and doesnot know much about php. So a little help from anyone is needed... Please help me out. My php script is shown below:
<?php
$con=mysqli_connect("localhost","username","password","db_name");
if (mysqli_connect_errno($con))
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$username = $_POST['username'];
$image = $_POST['image'];
$result = mysqli_query($con,"INSERT INTO table3 (username,imagename) VALUES ('$username','$image')");
if (!mysqli_query($con,$result))
{
die('Error: ' . mysqli_error($con));
}
else
echo "1 record added";
mysqli_close($con);
?>
1) **You have an error in your SQL syntax; ** means that You have error in your query. It seems that your query is okay but that error may come from your post data. you need to mysqli_real_escape_string for post data.
2) you have executed twice the query.
try like this :
$username = mysqli_real_escape_string($con, $_POST['username']);
$image = mysqli_real_escape_string($con,$_POST['image']);
$result = mysqli_query($con,"INSERT INTO table3 (username,imagename) VALUES ('$username','$image')");
if (!$result)
{
die('Error: ' . mysqli_error($con));
}
else
echo "1 record added";
mysqli_close($con);
You should have written something like
$result = mysqli_query($con,"INSERT INTO table3 (username,imagename) VALUES ('$username','$image')");
if (!$result)
{
die('Error: ' . mysqli_error($con));
}
the condition should be something like
if (!$result)
{
die('Error: ' . mysqli_error());
}
Your SQL request (INSERT) is sent at the 9th line and the result is caught in $result.
$result contains the number of lines affected by the previous request (1).
Then you call the mysqli_query method again with the value of $result as a SQL request : "1" is not a valid SQL request.

MySQL error on form submission

I'm getting a mysql error saying "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..."
Here's the basics of my code:
First I'm populating the select menu options with rows from the categories table. This is working fine:
<select id="dropdown-select" name="Name">
<option value="" id="dropdown-option">Please select a category.</option>
<?php
$query_categories = "SELECT * FROM categories";
$result_categories = mysql_query($query_categories) or die(mysql_error());
while($categories_row = mysql_fetch_array($result_categories)) {
echo '<option id="dropdown-option" value="' . $categories_row['cat_name'] . '">' . $categories_row['cat_name'] . '</option>';
}
?>
</select>
Later, when I go submit the form to the transactions table (the above table I pulled data from was the categories table, could this be a problem?) is when I get the error. I think its related to the above code bc if I remove this element from my form submission, it writes the rest of the values to the database without any errors.
if(!isset($_POST['Name'])) {
die('You must select an income or expense from the drop down menu.');
} else {
$Name = $_POST['Name'];
}
//create query
$query = "INSERT INTO transaction (month, trans_name, budgeted, actual) VALUES ('$Month', '$Name', $Budgeted', '$Actual')";
$result = mysql_query($query) or die("Error in query: $query. " . mysql_error());
Thanks for any help you can provide.
You are missing a single quote in your insert statement before $Budgeted
INSERT INTO transaction (month, trans_name, budgeted, actual) VALUES ('$Month', '$Name', '$Budgeted', '$Actual')"
If you have some fields which are defined in Database as VARCHAR, CHAR.
Also, if you are inserting a string value in Database from a PHP script, you need to add an enclosing single quote (') around it.
In your case, you are inserting a string without semicolons, so, it showing error in MySQL.
Your statement should be corrected by adding a single quote around $budget as:
$query = "INSERT INTO transaction (month, trans_name, budgeted, actual) VALUES ('$Month',
'$Name', '$Budgeted', '$Actual')";
------^
The error "You have an error in your SQL syntax" is exactly correct!
$query = "INSERT INTO transaction (month, trans_name, budgeted, actual)
VALUES ('$Month', '$Name', $Budgeted', '$Actual')";
Look here, you missed something ----^
There is a ' missing from your statement causing the syntax error. Put the single quote in and you should be good to go!

MySQL Syntax error on Insert Query from PHP

I'm getting a non-descriptive syntax error on a MYSQL query from PHP. If I "echo" the text of the query and paste it into a MySQL query window, the code works. Here is the SQL for the query, the error code, and the error message...
INSERT INTO ADVERTISEMENTS (`user_id`, `ad_name`, `click_url`, `img_url`, `bg_color`, `start_date`, `end_date`, `timer_delay`, `add_date`) VALUES (2, 'Test New Ad', 'http://www.google.com', 'red_arrow.png', '#000000', '1980-05-11 00:00:00', '2020-05-01 00:00:00', 5, '2013-07-14 22:21:59');
Error Code: 1064
Error Msg: 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 '' at line 1
Here is the PHP code I am using...
$link = mysqli_connect($UM_Settings["database_options"]["server_name"], $UM_Settings["database_options"]["username"], $UM_Settings["database_options"]["password"], $UM_Settings["database_options"]["database_name"]);
$advertisementNameNew = mysqli_real_escape_string($link, $_POST['advertisementNameNew']);
$destinationURLNew = mysqli_real_escape_string($link, $_POST['destinationURLNew']);
$dropboxUploadFile = mysqli_real_escape_string($link, $_POST['dropboxUploadFile']);
$backgroundColorNew = mysqli_real_escape_string($link, $_POST['backgroundColorNew']);
$bannerStartDateNew = DateStringToMySQL($_POST['bannerStartDateNew']);
$bannerEndDateNew = DateStringToMySQL($_POST['bannerEndDateNew']);
$bannerSetTimerNew = intval($_POST['bannerSetTimerNew']);
$tmpUserID = UM_GetCookie("UM_UserID");
$tmpAddDate = DateStringToMySQL('now');
echo "INSERT INTO ADVERTISEMENTS(`user_id`, `ad_name`, `click_url`, `img_url`, `bg_color`, `start_date`, `end_date`, `timer_delay`, `add_date`) VALUES ($tmpUserID, '$advertisementNameNew', '$destinationURLNew', '$dropboxUploadFile', '$backgroundColorNew', '$bannerStartDateNew', '$bannerEndDateNew', $bannerSetTimerNew, '$tmpAddDate');<br />";
if (!mysqli_query($link, "INSERT INTO ADVERTISEMENTS(`user_id`, `ad_name`, `click_url`, `img_url`, `bg_color`, `start_date`, `end_date`, `timer_delay`, `add_date`) VALUES ($tmpUserID, '$advertisementNameNew', '$destinationURLNew', '$dropboxUploadFile', '$backgroundColorNew', '$bannerStartDateNew', '$bannerEndDateNew', $bannerSetTimerNew, '$tmpAddDate');")) {
printf("Error Code: %s\n", mysqli_errno($link));
echo "<br />";
printf("Error Msg: %s\n", mysqli_error($link));
}
I know that the database connection is working. I am able to select and update tables. I can also insert into other tables with different queries.
I am open to any suggestions.
Thank you in advance for your help!
I see a few errors in your query strings.
First, all your variables are passed as literal strings: "... VALUES ($tmpUserID, '$advertisementNameNew', ..." should be "... VALUES (".$tmpUserID.", '".$advertisementNameNew."', ...".
Second, I see missing quotes around $bannerSetTimerNew.
Third, there is an extra ;.
here's how I would write the query:
if (!mysqli_query($link, "INSERT INTO ADVERTISEMENTS (user_id, ad_name, click_url, img_url, bg_color, start_date, end_date, timer_delay, add_date) VALUES (".$tmpUserID.", '".$advertisementNameNew."', '".$destinationURLNew."', '".$dropboxUploadFile."', '".$backgroundColorNew."', '".$bannerStartDateNew."', '".$bannerEndDateNew."', '".$bannerSetTimerNew."', '".$tmpAddDate."')")) { ...
I didnt test it though.
hope this helps.
I see a ; at the end of the query. Are you sure that should be there?
There are two things
1. Remove the ; from at the end of the query.
2. I hope timer_delay field has datatype "Int" if its a VARCHAR then you will have to include quotes for that field value.
I hope this will help.
Passerby, thank you for your comment. This was my first experience with using mysqli, I changed my query to use the "bind_param" method, and everything works now. For anyone else with a similar problem, here is the corrected code...
$mysqli = new mysqli($UM_Settings["database_options"]["server_name"], $UM_Settings["database_options"]["username"], $UM_Settings["database_options"]["password"], $UM_Settings["database_options"]["database_name"]);
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}
$advertisementNameNew = $_POST['advertisementNameNew'];
$destinationURLNew = $_POST['destinationURLNew'];
$dropboxUploadFile = $_POST['dropboxUploadFile'];
$backgroundColorNew = $_POST['backgroundColorNew'];
$bannerStartDateNew = DateStringToMySQL($_POST['bannerStartDateNew']);
$bannerEndDateNew = DateStringToMySQL($_POST['bannerEndDateNew']);
$bannerSetTimerNew = intval($_POST['bannerSetTimerNew']);
$tmpUserID = UM_GetCookie("UM_UserID");
$tmpAddDate = DateStringToMySQL('now');
/* Prepared statement, stage 1: prepare */
if (!($stmt = $mysqli->prepare("INSERT INTO `ADVERTISEMENTS` (`user_id`, `ad_name`, `click_url`, `img_url`, `bg_color`, `start_date`, `end_date`, `timer_delay`, `add_date`) VALUES (?,?,?,?,?,?,?,?,?)"))) {
echo "Prepare failed: (" . $mysqli->errno . ") " . $mysqli->error;
}
if (!$stmt->bind_param("issssssis",$tmpUserID, $advertisementNameNew, $destinationURLNew, $dropboxUploadFile, $backgroundColorNew, $bannerStartDateNew, $bannerEndDateNew, $bannerSetTimerNew, $tmpAddDate)) {
echo "Binding parameters failed: (" . $stmt->errno . ") " . $stmt->error;
}
if (!$stmt->execute()) {
echo "Execute failed: (" . $stmt->errno . ") " . $stmt->error;
}
$_GET['ad_id'] = $stmt->insert_id;
$stmt->close();

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