CODEWITHSUNDEEP
phpmysqlsql
Lim QingYong
Lim QingYong

Reputation: 3

Database not inserting

projectcostingfile table I have checked my sql query several times and make sure that the names are correct. Still not inserting.

if(isset($_POST['submit'])){//to run PHP script on submit
if(!empty($_POST['check_list'])){
// Loop to store and display values of individual checked checkbox.
foreach($_POST['check_list'] as $selected){
echo $selected."</br>";
$sqlinsert="INSERT INTO projectcostingfile (material) VALUES ('".$selected."')";
mysqli_query($conn, $sqlinsert);
}
}
}

The variable $selected is echoed correctly. There is no error when I run the code too. Any help will be appreciated!

Thanks for your help. The problem lies with my table structure like what some of you mentioned. I set the column fields to NOT NULL. Therefore inserting into only 1 column would not have any INSERT record shown in the database.

Upvotes: 0

Views: 83

Answers (7)

Pavan Baddi
Pavan Baddi

Reputation: 479

Most of the solutions have been given above. But i would like to tell you that even a faced such issue were my code, Mysql connnection, query , php codes Where right but in myphpadmin inside my database i have accidentally added a 'black Space' near sql table or column name because to which my sql query was not giving me the error but also not inserting the code.i recommend you to try this.

If doesn't work than my bad.

Upvotes: 0

bobjan
bobjan

Reputation: 363

Definitely there is nothing wrong with your PHP code, but it does not comply DB schema. Check error.log file on your PHP server and it will show you lot of DB errors, concerning NULL values.

Upvotes: 0

Ganesh Radhakrishnan
Ganesh Radhakrishnan

Reputation: 350

I think you cannot insert only one field material in your given table. From the picture shown in your question, all fields in your table should not be null. So i think we cannot added rows in that table by filling only one field. You have to insert all the fields or change the null status to yes of all other field except material.

I think this might work if you don't have any other options.

Upvotes: 0

Prateek Verma
Prateek Verma

Reputation: 889

Please try this:

if(isset($_POST['submit'])){//to run PHP script on submit
if(!empty($_POST['check_list'])){
// Loop to store and display values of individual checked checkbox.
$check_values = implode(",",$_POST['check_list']);
echo $check_values;
$sqlinsert="INSERT INTO projectcostingfile (material) VALUES ('".$check_values."')";
mysqli_query($conn, $sqlinsert);

}
}

Hope, this will help you.

Upvotes: 0

VK.Anup
VK.Anup

Reputation: 33

if(isset($_POST['submit'])){//to run PHP script on submit
if(!empty($_POST['check_list'])){
// Loop to store and display values of individual checked checkbox.
foreach($_POST['check_list'] as $selected){
echo $selected."</br>";
$sqlinsert="INSERT INTO projectcostingfile (material) VALUES ('".$selected."')";
mysqli_query($conn, $sqlinsert);
}
}
}

above code is your correct you just put following code such as post array

<input type="text" name="check_list[]">

Upvotes: 0

Kashan Shah
Kashan Shah

Reputation: 138

Use 

$sqlinsert="INSERT INTO projectcostingfile SET material = '".$selected."'";

Also use 'or die(mysqli_error());' after mysqli_query to identify the error..

mysqli_query($conn, $sqlinsert) or die(mysqli_error($conn));

Upvotes: 3

Sujal Patel
Sujal Patel

Reputation: 2532

Try to Replace Below Syntax : 

$sqlinsert="INSERT INTO projectcostingfile (`material`) VALUES ('".$selected."')";

Upvotes: 0

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