CODEWITHSUNDEEP
phpmysql
Marlon
Marlon

Reputation: 111

Issue trying to insert data into a MySQL table with php

I have an existing table and I'm trying to insert data to it from a form submission.

I think I need another set of eyes to look over my code. There is an issue with it, at this point I'm not sure what.

I fill out the form, it appears to submit, I check the table and I get nothing. No data gets inserted.

If some of the values aren't being used right now, can that hinder the data from being inserted into the table? Shouldn't the form inputs that are on the form be inserted anyway? I tried to use only the values within the form, and that still did not work.

For example, right now the form inputs are only for first, last, title, genre, about, and picture. The rest only exist within the table. Data for the remaining fields cannot be currently entered at this time.

Hopefully someone has a solution?

$servername = "server";
$username = "user";
$password = "pass";
$dbname = "db";

if (isset($_POST['submit'])){
$conn = mysqli_connect($servername,$username,$password,$dbname);
if (!$conn) {
    die("Connection failed: " . mysqli_error());
}


$first      = mysqli_real_escape_string($conn, $_POST['First']);
$last       = mysqli_real_escape_string($conn, $_POST['Last']);
$title      = mysqli_real_escape_string($conn, $_POST['Title']);
$storylink  = mysqli_real_escape_string($conn, $_POST['StoryLink']);
$genre      = mysqli_real_escape_string($conn, $_POST['Genre']);
$about      = mysqli_real_escape_string($conn, $_POST['About']);
$link       = mysqli_real_escape_string($conn, $_POST['Link']);
$picture    = mysqli_real_escape_string($conn, $_POST['Picture']);
$alt        = mysqli_real_escape_string($conn, $_POST['ALT']);

$sql = "INSERT INTO ContrTemp (`First`,`Last`,`Title`,`StoryLink`,`Genre`,`About`,`Link`,`Picture`,`ALT`)
VALUES ('$first','$last','$title','$storylink','$genre','$about','$link','$picture','$alt')";

mysqli_query($conn, $sql) or die('Error: ' . mysqli_error($conn));
mysqli_close($conn);
}

Here is one input field from the form. The others are pretty much the same.

<input type="text" id="ContrTitle" name="Title" placeholder="Title" class="inputFields" style="width:650px;" />

Could there be an issue with the name within the input?

sql table structure:

CREATE TABLE IF NOT EXISTS `ContrTemp` (
  `ID` int(5) NOT NULL AUTO_INCREMENT,
  `First` varchar(40) COLLATE utf8_unicode_ci NOT NULL,
  `Last` varchar(40) COLLATE utf8_unicode_ci NOT NULL,
  `Title` varchar(50) COLLATE utf8_unicode_ci NOT NULL,
  `StoryLink` varchar(140) COLLATE utf8_unicode_ci NOT NULL,
  `Genre` varchar(11) COLLATE utf8_unicode_ci NOT NULL,
  `About` varchar(2000) COLLATE utf8_unicode_ci NOT NULL,
  `Link` varchar(125) COLLATE utf8_unicode_ci NOT NULL,
  `Picture` varchar(500) COLLATE utf8_unicode_ci NOT NULL,
  `ALT` varchar(100) COLLATE utf8_unicode_ci NOT NULL,
  PRIMARY KEY (`ID`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_unicode_ci AUTO_INCREMENT=1 ;

Now I'm actually hitting the database, a record shows up, but no data was inserted. Before, No record was inserted at all.

Upvotes: 1

Views: 121

Answers (4)

Marlon
Marlon

Reputation: 111

I had hoped this had been as simple as the code I provided above, thus the reason I did not mention that this code was being used as part of a WordPress template. I did not think that would be come an issue as the template is pretty light weight.

With that said, I simply took my block of php code that handles the data insertion, and I placed it at the top of the template. Worked like a charm..

Upvotes: 0

Martin
Martin

Reputation: 22760

Your mysqli_error() is incorrect syntax, you need to specify the connection so have mysqli_error($conn) to properly feedback to you the SQL errors. As mentioned by Mitkosoft in comments some of your column names are MySQL keywords. It could be a good habit to get into to always encase column names in backticks.

So:

$sql = "INSERT INTO ContrTemp (`First`,`Last`,`Title`,`StoryLink`,`Genre`,`About`,`Link`,`Picture`,`ALT`)
VALUES ('$first','$last','$title','$storylink','$genre','$about','$link','$picture','$alt')";

mysqli_query($conn, $sql) or die('Error: ' . mysqli_error($conn));

Also you don't need to select db with mysqli_select_db($conn,$dbname); because the database is selected on the login command mysqli_connect above.

Further queries:

  • Is your database access account allowed to INSERT to this database?
  • Can you insert the same data as in your form directly into the database using an interface such as PHPMyAdmin, MySQL Workbench (Or whichever one you use to view the DB)?

Upvotes: 1

Sanooj T
Sanooj T

Reputation: 1327

I think you have done pretty well but one last try put these symbols [backticks] ` around your fields to understand mysql they are just fields.

$sql = "INSERT INTO ContrTemp (`First`,`Last`,`Title`,`StoryLink`,`Genre`,`About`,`Link`,`Picture`,`ALT`)
VALUES ('$first','$last','$title','$storylink','$genre','$about','$link','$picture','$alt')";

Upvotes: 2

ram singh
ram singh

Reputation: 140

You can try this one for inserting data 

$sql = 'INSERT INTO ContrTemp  (First,Last,Title,StoryLink,Genre,About,Link,Picture,ALT) VALUES ("'.$first.'","'.$last.'","'.$title.'","'.$storylink.'","'.$genre.'","'.$about.'","'.$link.'","'.$picture.'","'.$alt.'")';

Upvotes: 2

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