mysql insert error check - php

I have a PHP script that collects form data and inserts some of that data into a MySQL database. I just noticed that some inserts/records were NOT, or never created in the database. I would like to write a retry routine that if the insert fails to retry 3 times and then error out to the user.
Just so you can see my code for the DB and the insert so you can see that I am NOT nuts...
mysql_connect($hostname,$username, $password) or die(mysql_error());
mysql_select_db($dbname) or die(mysql_error());
Here is my insert code:
$query = "INSERT INTO contacts VALUES ('','$name','$address','$city','$state','$zip','$phone','$email_address','$arrive','$depart','$room','$found','$promocode','$message','$datetimestamp','$ip')";
mysql_query($query);
mysql_close();
I started out with an IF statement then into a loop but got lost.
#Jay:
So something like this:
$conn = new mysqli($hostname, $username, $password, $dbname);
// check connection
if (mysqli_connect_errno())
{
exit('Connect failed: '. mysqli_connect_error());
}
$query = "INSERT INTO contacts VALUES ('','$name','$address','$city','$state','$zip','$phone','$email_address','$arrive','$depart','$room','$found','$promocode','$message','$datetimestamp','$ip')";
// Performs the $query on the server to insert the values
if ($conn->query($query) === TRUE) {
//echo 'users entry saved successfully';
}
else {
echo 'Error: '. $conn->error;
}
$conn->close();
I am already checking for injection before the insert query

Your query is perfect, make sure that number of parameter you are passing in sql query is same as number of column in database table & parameter value in sql is same order of database table column order

Related

Inconsistency of MySQL syntax errors depending on whether query made from command line or mysqli_connect

A simple MySQL query is returning a syntax error over mysqli_connect, but the identical, copy-pasted query is successful in both the CLI and phpMyAdmin.
Consider this example for MySQL 8.0:
// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$sql = "USE aTable; INSERT INTO aTable (`aColumn`) VALUES ('aValue');";
if (mysqli_query($conn, $sql)) {
echo "New record created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
When the PHP runs it prints an error:
USE aTable; INSERT INTO aTable (aColumn) VALUES ('aValue');
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 'INSERT INTO aTable (aColumn) VALUES ('aValue')' at line 1
However when the same query is pasted into phpMyAdmin it tells me:
1 row inserted.
Inserted row id: 6 (Query took 0.0063 seconds.)
USE aTable; INSERT INTO aTable (aColumn) VALUES ('aValue');
Why are they different?
Remove the USE from the PHP code because the database is already selected in the mysqli_connect method.
$sql = "USE aTable; INSERT INTO aTable (`aColumn`) VALUES ('aValue');";
Should be:
$sql = "INSERT INTO aTable (`aColumn`) VALUES ('aValue');";
Also, make sure that you're not using the database name in the INSERT query.

Escape strings/ inserting in php script

I'm trying to finish a script that connects to two databases, each on a different server, and preforms an update. Basically, the tables being selected from and inserted to are identical: I did a dump/import the other day. The script needs to keep my local table up to date from the remote once since there will be daily records inserted into the remote one and I need to keep it up to date locally.
The key here is that I'm determining the new rows on the remote server by looking at the Auto-incremented Primary key that the tables share, SESSIONID . I'm trying to get my loop below to say, if the id exists in remote server and not local, then insert those records in local server.
I run the below script in powershell by typing php 'filename', and I get both of my successful connection messages, and then I get this message: Incorrect datetime value: '' for column 'ALERTINGTIMESTAMP' at row 1. In this record it's trying to insert, the datetime value is NULL, which the table allows for, however I'm worried it's an issue with escaping characters or something.
How can I modify this to escape properly, or get these records inserted.
Note: Replication and large dump/import/table recreations are not an option for us in this situation. We have several similar scripts to this running and we want to keep the same process here. I'm merely looking to resolve these errors or have someone give me a more efficient way of coding this script, perhaps using a max id or something along those lines.
Here's the script:
ini_set('memory_limit', '256M');
// Create connection
$conn = new mysqli($servername, $username, $password);
$conn2 = new mysqli($servername2, $username2, $password2);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
// Check connection2
if ($conn2->connect_error) {
die("Connection failed: " . $conn2->connect_error);
}
echo "Connected successfully";
$latest_result = $conn2->query("SELECT MAX(`SESSIONID`) FROM `ambition`.`session`");
$latest_row = $latest_result->fetch_row();
$latest_session_id = $latest_row[0];
//Select All rows from the source phone database
$source_data = mysqli_query($conn, "SELECT * FROM `cdrdb`.`session` WHERE `SESSIONID` > $latest_session_id");
// Loop on the results
while($source_item = $source_data->fetch_assoc()) {
// Check if row exists in destination phone database
$row_exists = $conn2->query("SELECT SESSIONID FROM ambition.session WHERE SESSIONID = '".$source_item['SESSIONID']."' ") or die(mysqli_error($conn2));
//if query returns false, rows don't exist with that new ID.
if ($row_exists->num_rows == 0){
//Insert new rows into ambition.session
$conn2->query("INSERT INTO ambition.session (SESSIONID,SESSIONTYPE,CALLINGPARTYNO,FINALLYCALLEDPARTYNO,DIALPLANNAME,TERMINATIONREASONCODE,ISCLEARINGLEGORIGINATING,CREATIONTIMESTAMP,ALERTINGTIMESTAMP,CONNECTTIMESTAMP,DISCONNECTTIMESTAMP,HOLDTIMESECS,LEGTYPE1,LEGTYPE2,INTERNALPARTYTYPE1,INTERNALPARTYTYPE2,SERVICETYPEID1,SERVICETYPEID2,EXTENSIONID1,EXTENSIONID2,LOCATION1,LOCATION2,TRUNKGROUPNAME1,TRUNKGROUPNAME2,SESSIONIDTRANSFEREDFROM,SESSIONIDTRANSFEREDTO,ISTRANSFERINITIATEDBYLEG1,SERVICEEXTENSION1,SERVICEEXTENSION2,SERVICENAME1,SERVICENAME2,MISSEDUSERID2,ISEMERGENCYCALL,NOTABLECALLID,RESPONSIBLEUSEREXTENSIONID,ORIGINALLYCALLEDPARTYNO,ACCOUNTCODE,ACCOUNTCLIENT,ORIGINATINGLEGID,SYSTEMRESTARTNO,PATTERN,HOLDCOUNT,AUXSESSIONTYPE,DEVICEID1,DEVICEID2,ISLEG1ORIGINATING,ISLEG2ORIGINATING,GLOBALCALLID,CADTEMPLATEID,CADTEMPLATEID2,ts,INITIATOR,ACCOUNTNAME,APPNAME,CALLID,CHRTYPE,CALLERNAME,serviceid1,serviceid2)
VALUES ('".$source['SESSIONID']."' ,
'".$source['SESSIONTYPE']."' ,
'".$source['CALLINGPARTYNO']."' ,
'".$source['FINALLYCALLEDPARTYNO']."',
'".$source['DIALPLANNAME']."',
'".$source['TERMINATIONREASONCODE']."',
'".$source['ISCLEARINGLEGORIGINATING']."',
'".$source['CREATIONTIMESTAMP']."',
'".$source['ALERTINGTIMESTAMP']."',
'".$source['CONNECTTIMESTAMP']."',
'".$source['DISCONNECTTIMESTAMP']."',
'".$source['HOLDTIMESECS']."',
'".$source['LEGTYPE1']."',
'".$source['LEGTYPE2']."',
'".$source['INTERNALPARTYTYPE1']."',
'".$source['INTERNALPARTYTYPE2']."',
'".$source['SERVICETYPEID1']."',
'".$source['SERVICETYPEID2']."',
'".$source['EXTENSIONID1']."',
'".$source['EXTENSIONID2']."',
'".$source['LOCATION1']."',
'".$source['LOCATION2']."',
'".$source['TRUNKGROUPNAME1']."',
'".$source['TRUNKGROUPNAME2']."',
'".$source['SESSIONIDTRANSFEREDFROM']."',
'".$source['SESSIONIDTRANSFEREDTO']."',
'".$source['ISTRANSFERINITIATEDBYLEG1']."',
'".$source['SERVICEEXTENSION1']."',
'".$source['SERVICEEXTENSION2']."',
'".$source['SERVICENAME1']."',
'".$source['SERVICENAME2']."',
'".$source['MISSEDUSERID2']."',
'".$source['ISEMERGENCYCALL']."',
'".$source['NOTABLECALLID']."',
'".$source['RESPONSIBLEUSEREXTENSIONID']."',
'".$source['ORIGINALLYCALLEDPARTYNO']."',
'".$source['ACCOUNTCODE']."',
'".$source['ACCOUNTCLIENT']."',
'".$source['ORIGINATINGLEGID']."',
'".$source['SYSTEMRESTARTNO']."',
'".$source['PATTERN']."',
'".$source['HOLDCOUNT']."',
'".$source['AUXSESSIONTYPE']."',
'".$source['DEVICEID1']."',
'".$source['DEVICEID2']."',
'".$source['ISLEG1ORIGINATING']."',
'".$source['ISLEG2ORIGINATING']."',
'".$source['GLOBALCALLID']."',
'".$source['CADTEMPLATEID']."',
'".$source['CADTEMPLATEID2']."',
'".$source['ts']."',
'".$source['INITIATOR']."',
'".$source['ACCOUNTNAME']."',
'".$source['APPNAME']."',
'".$source['CALLID']."',
'".$source['CHRTYPE']."',
'".$source['CALLERNAME']."',
'".$source['serviceid1']."',
'".$source['serviceid2']."')");
}
}
Like Pankaj said, try something like this:
//Insert new rows into ambition.session
$statement = $conn2->prepare('INSERT INTO ambition.session (SESSIONID,SESSIONTYPE,CALLINGPARTYNO,FINALLYCALLEDPARTYNO,DIALPLANNAME,TERMINATIONREASONCODE,ISCLEARINGLEGORIGINATING,CREATIONTIMESTAMP,ALERTINGTIMESTAMP,CONNECTTIMESTAMP,DISCONNECTTIMESTAMP,HOLDTIMESECS,LEGTYPE1,LEGTYPE2,INTERNALPARTYTYPE1,INTERNALPARTYTYPE2,SERVICETYPEID1,SERVICETYPEID2,EXTENSIONID1,EXTENSIONID2,LOCATION1,LOCATION2,TRUNKGROUPNAME1,TRUNKGROUPNAME2,SESSIONIDTRANSFEREDFROM,SESSIONIDTRANSFEREDTO,ISTRANSFERINITIATEDBYLEG1,SERVICEEXTENSION1,SERVICEEXTENSION2,SERVICENAME1,SERVICENAME2,MISSEDUSERID2,ISEMERGENCYCALL,NOTABLECALLID,RESPONSIBLEUSEREXTENSIONID,ORIGINALLYCALLEDPARTYNO,ACCOUNTCODE,ACCOUNTCLIENT,ORIGINATINGLEGID,SYSTEMRESTARTNO,PATTERN,HOLDCOUNT,AUXSESSIONTYPE,DEVICEID1,DEVICEID2,ISLEG1ORIGINATING,ISLEG2ORIGINATING,GLOBALCALLID,CADTEMPLATEID,CADTEMPLATEID2,ts,INITIATOR,ACCOUNTNAME,APPNAME,CALLID,CHRTYPE,CALLERNAME,serviceid1,serviceid2)
VALUES (?, ?, ?, ...);');
$statement->bindParam(1, $source['SESSIONID']);
$statement->bindParam(2, $source['SESSIONTYPE']);
$statement->bindParam(3, $source['CALLINGPARTYNO']);
//...
$statement->execute();
You have to use prepare() function to use parameterized query. Here I have taken example of your query with few parameters you can add yourself with other variables.
$stmt = $conn2->prepare("INSERT INTO ambition.session (SESSIONID,SESSIONTYPE,CALLINGPARTYNO,FINALLYCALLEDPARTYNO) VALUES (:SESSIONID ,:SESSIONTYPE ,:CALLINGPARTYNO ,:FINALLYCALLEDPARTYNO)");
$stmt->bindParam(':SESSIONID', $source['SESSIONID']);
$stmt->bindParam(':SESSIONTYPE', $source['SESSIONTYPE']);
$stmt->bindParam(':CALLINGPARTYNO', $source['CALLINGPARTYNO']);
$stmt->bindParam(':FINALLYCALLEDPARTYNO', $source['FINALLYCALLEDPARTYNO']);
$stmt->execute();
You can checkout this link for more understanding. http://php.net/manual/en/mysqli.prepare.php

How can I get mysql to print rows from a database table

I am trying to learn php from W3schools which includes a mysql section.So far I have completed every other part of the tutorial on w3school except the part that prints content from a database table. For some very weird reason , nothing displays when I run my code. Please how can I get this working and could my problem come from the fact that I am using MariaDB with Xampp instead of Mysql although they said it was practically the same syntax.
Here is the code
<?php
$servername = "localhost";
$username = "uhexos";
$password = "strongpassword";
$database = "fruitdb";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// Create database
$sql = "CREATE DATABASE fruitDB";
if ($conn->query($sql) === TRUE) {
echo "Database created successfully";
} else {
echo "Error creating database: " . $conn->error;
}
$conn->close();
// Create connection
$conn = mysqli_connect($servername, $username, $password,$database);
// sql to create table
$complexquery = "CREATE TABLE MyFruits (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
FruitType VARCHAR(30) NOT NULL,
FruitTaste VARCHAR(30) NOT NULL,
FruitQuantity INT NOT NULL,
DatePurchased TIMESTAMP
)";
if ($conn->query($complexquery) === TRUE) {
echo "Table Fruits created successfully<br> ";
} else {
echo "Error creating table: " . $conn->error;
}
$entry = "INSERT INTO myfruits (fruittype,fruittaste,fruitquantity) VALUES ('orange','sweet','50'),('lemon','sour','10'),('banana','sweet','15')";
if ($conn->query($entry) === TRUE) {
echo "New records created successfully";
} else {
echo "Error: " . $conn->error;
}
$sql = 'SELECT id, fruitname, fruittaste FROM myfruits';
$retval = mysql_query( $sql, $conn );
if(! $retval ) {
die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) {
echo "EMP ID :{$row['id']} <br> ".
"EMP NAME : {$row['fruitname']} <br> ".
"EMP SALARY : {$row['fruittaste']} <br> ".
"--------------------------------<br>";
}
echo "Fetched data successfully\n";
mysql_close($conn);
?>
this is the output I get from all my echos.
Error creating database: Can't create database 'fruitdb'; database existsError creating table: Table 'myfruits' already existsNew records created successfully
or
Database created successfullyTable Fruits created successfully
New records created successfully
Based on the error message, you managed to create the database and tables once and now each time you run the code it fails because you can't reuse the names.
You definitely don't want to have code trying to erase & start fresh on your database every time. In fact, most often I find that you don't even create the database inside your regular code but use phpMyAdmin or some other admin page to do that. But creating tables inside code is normal enough. Two options:
1 - Create the table only if it does not already exist. This is extremely safe. However, if you want to start a table over again with a new structure, or start with it always empty, that won't work. To do that, just change CREATE TABLE to CREATE TABLE IF NOT EXISTS
2 - Delete the table before creating it. Before each CREATE TABLE command, add a command like DELETE TABLE IF EXISTS MyFruits
Remember database name is Case-insensitive, so it doesn't matter whether you create a DB name "fruitdb" or "fruitDb" both are same.That is the reason you are getting error. Also you don't have to create a new database when you execute any file. If you have already created the database than you only have make the connection with it.
Let's debug your code line by line.
Line 8 -
// Create connection
$conn = new mysqli($servername, $username, $password);
Here you are creating the connection with your database because you have already created that database. If you check your phpmyadmin, you'll find a database named "fruitdb"
Line 10 -
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
Here your checking whether the you are able to connect with your database. If not it will throw the error and your script will stop. Right now your code successfully runs till this point.
Line 15 -
// Create database
$sql = "CREATE DATABASE fruitDB";
Here you are again creating a database with same name and your code stops working as you already have it.
The error was from this line
$sql = 'SELECT id, fruitname, fruittaste FROM myfruits';
I accidentally put fruitname instead of fruittype and that is what caused it to fail. So for anyone else with thi problem my advice is to check your variable names if you are 100% sure of your syntax. Thanks for all the help.

I want to insert a value from an array into a column of sql table, but the content is not going into the table?

for($i=0;$i<=feof($getdata);$i++)
{
if (filter_var($data[$i][1], FILTER_VALIDATE_EMAIL)){
echo $data[$i][1];
$email=$data[$i][1];
$conn = mysqli_connect($dbhost,$dbuser,$dbpass, $dbname);
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql ="INSERT INTO promo_user (uid,name,email) VALUES (,'', '$email')";
mysqli_query($sql,$conn);
mysqli_close($conn);
i am using the above code but there is something wrong with it,whenever i run the code the echo is working fine but the content does go into sql table
Please help
You have the arguments transposed - the correct order is
mysqli_query($con,$query)
So the first parameter of mysqli_query will be the connection and second is query.
Also, you can make connection outside the loop, so connection will be initiated only one time.

im getting error when performing insert multiple records

<?php
session_start();
if(isset($_SESSION['id']))
$servername = "localhost";
$dbname = "school";
// Create connection
$conn = mysqli_connect($servername, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
$teacher_id=$_SESSION['id'];
$student_id=$_POST['student_id'];
$subject=$_POST['subject'];
$description=$_POST['description'];
$sql="INSERT INTO t_sent (teacher_id, student_id, subject, description)
VALUES
('$_SESSION[id]', '$_POST[student_id]', '$_POST[subject]', '$_POST[description]')";
$sql .="INSERT INTO p_inbox (teacher_id, student_id, subject, description)
VALUES
('$_SESSION[id]', '$_POST[student_id]', '$_POST[subject]', '$_POST[description]')";
if (mysqli_multi_query($conn, $sql)) {
echo "New records created successfully";
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}
mysqli_close($conn);
?>
and im getting this error message when im adding records
Error:
INSERT INTO t_sent (teacher_id, student_id, subject, description) VALUES ('Badri', 'ca11099', 'cm ', 'cm')INSERT INTO p_inbox (teacher_id, student_id, subject, description) VALUES ('Badri', 'ca11099', 'cm ', 'cm')
No database selected
I don't know what I'm missing
I think you missed out the syntax for mysqli_connet
mysqli_connect(host, user,password, db);
Please let me know if it didnt work for you.
First of all it looks like you are incorrectly using an if statement - namely
if(isset($_SESSION['id'])).
That if statement will only apply to the very next line ($servername = "localhost";). You need to wrap the code you want that statement to apply to with { } (judging by your code, it's every line after that).
Your parameters for mysqli_connect() are wrong - it should be
mysqli_connect($servername, $username, $password, $dbname);
You are missing the username and password which should go before the database name.
You're also not sanitizing or escaping your input which is very bad and can lead to SQL injection attacks. You should never directly take user input and run queries against that.
Use correct mysqli_connect() with all the parameters:
<?php
$con = mysqli_connect($servername, $username,$password, $database_name);
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
?>
Also, You can select database using mysqli_select_db() function

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