PHP insert statement returns no errors but does not insert

Question

I have two very long list:

$countryCode

And

$countryName

To insert these list i have created the following loop:

   $i = 0;

$con=mysqli_connect("localhost","root","root","system_bloglic_com_test");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

foreach($countryCode as $cCode){
mysqli_query($con,"INSERT INTO Contries (contries_id, contries_name, contries_code)
VALUES(NULL,$cCode,$countryName[$i])");
$i++;
}

When i run this i get no erros but no rows have been inserted into database how come?

You won't get errors because you aren't checking for any errors. — Ben Fortune
– Ben Fortune, Commented Aug 20, 2013 at 13:50
What @BenFortune said, also do you mean to be typing "countries" as "contries" interchangeably within the query? — ಠ_ಠ
– ಠ_ಠ, Commented Aug 20, 2013 at 13:50
country name and code are string so you need to insert them as strings: — AlecTMH
– AlecTMH, Commented Aug 20, 2013 at 13:51

Ahmed Alaa El-Din · Accepted Answer · 2013-08-20 13:52:02Z

2

did you check mysqli_error(); it will help you to find exactly the error. It returns FALSE if an error occurred. In that case mysqli_error() gives you more information about the error.

$result = mysqli_query($con,"INSERT INTO Contries (contries_id, contries_name, contries_code)
VALUES(NULL,$cCode,$countryName[$i])");
if ( false===$result ) {
  printf("error: %s\n", mysqli_error($con));
}
else {
  echo 'done.';
}

answered Aug 20, 2013 at 13:52

Ahmed Alaa El-Din

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Comments

Ghazanfar Mir · Accepted Answer · 2013-08-20 14:18:08Z

1

Your string values should always be enclosed within quotes

mysqli_query($con,"INSERT INTO Contries (contries_id, contries_name, contries_code)
VALUES(NULL,'{$cCode}','{$countryName[$i]}')");

and also you should use error handling as instructed by others using die() method

edited Aug 20, 2013 at 14:18

answered Aug 20, 2013 at 14:11

Ghazanfar Mir

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Comments

nl-x · Accepted Answer · 2013-08-20 13:53:30Z

1

You should change

mysqli_query(...);

into

mysqli_query(...) or die("error in query: " . mysqli_error($con));

To echo out any errors. Otherwise errors are silently ignored.

answered Aug 20, 2013 at 13:53

nl-x

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Comments

Ben Fortune · Accepted Answer · 2013-08-20 15:02:44Z

0

You need to check for errors in your code.

if(!mysqli_query($con,"INSERT INTO Contries (contries_id, contries_name, contries_code)
VALUES(NULL,'$cCode','$countryName[$i]')")){
    echo mysqli_error($con);
}

edited Aug 20, 2013 at 15:02

answered Aug 20, 2013 at 13:52

Ben Fortune

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