通过PHP将数据从多语言HTML表单插入MySQL表

Question

I have HTML form as below:

For now there are 3 languages in future will be more, fields with prefix _lang are generating dynamically.

These are input fields ( name_en , name_ru , etc. are placeholders).

MySQL tables looks like:

Parent table

Id    UserId

Details table

Id    ParentId    Name     Description    Language

I would like to insert data to Details like this:

Id    ParentId    Name     Description    Language
1     1           FooEn    FooBarEn       EN
2     1           FooRu    FooBarRu       RU
3     1           FooFr    FooBarFr       FR

And to Parent table like this:

Id    UserId
1     123

After form submitted I can insert successfully data to Parent table. I've faced a problem with inserting this data to Details table. I don't have idea how could I merge name_en and description_en as 1 entry.

In PHP I have $language_array with all possible languages (for this example en , ru , fr ) also I'm using substr to get language code from data, but what next? Should I use foreach somehow?

foreach($language_array as $k=>$v) {
   // there I've missed my mind
}

First time I've faced HTML form structure like this. Please, give me any suggestion.

Answer 1

The simplest solution would be to retrieve the form fields from your $_POST data.

Assuming your $_POST data has the following structure.

Array
(
    [name_en] => name_en
    [name_ru] => name_ru
    [name_fr] => name_fr
    [description_en] => description_en
    [description_ru] => description_ru
    [description_fr] => description_fr
)

You can then use your $language_array to iterate over and extract the associated $_POST form field values like so: $_POST['fieldname_' . $lang] $_POST['fieldname_' . $lang]

$languages = ['en', 'fr', 'ru'];
$sql = 'INSERT INTO `table_name` (`Name`, `Description`, `Language`) VALUES (?, ?, ?)';
if ($stmt = mysqli_prepare($conn, $sql)) {
    mysqli_stmt_bind_param($stmt, 'sss', $name, $description, $language);
    foreach ($languages as $lang) {
       $name = array_key_exists('name_' . $lang, $_POST) ? $_POST['name_' . $lang] : null;
       $description = array_key_exists('description_' . $lang, $_POST) ? $_POST['description_' . $lang] : null;
       $language = strtoupper($lang);
       mysqli_stmt_execute($stmt);
    }
}

For details on mysqli and usage on prepared statements, please see the PHP documentation .

---

An alternate solution would be to create a listing of form fields that you want to have displayed to the user.

$fields = ['name' => 'Name', 'description' => 'Description'];
$languages = ['en', 'ru', 'fr'];

Then you can iterate over each field and language to render them in the browser.

Take note that the input field names are prefixed with their associated language as an array of fields, such as LANG[field]

foreach ($fields as $fieldName => $section) {
    echo '<fieldset>';
    echo '<legend>' . $section. '</legend>';
    foreach ($languages as $language) {
        $inputName = strtoupper($language) . '[' . $fieldName . ']';
        $placeholder = $fieldName  . '_' . $language;
        echo '<p>';
        echo '<label>';
        echo '<input type="text" name="' . $inputName . '" placeholder="' . $placeholder . '"/>';
        echo '</label>';
        echo '</p>';
    }
    echo '</fieldset>';
}

Resulting Form:

Your $_POST data would then have the following structure

Array
(
    [EN] => Array
        (
            [name] => name_en
            [description] => description_en
        )

    [RU] => Array
        (
            [name] => name_ru
            [description] => description_ru
        )

    [FR] => Array
        (
            [name] => name_fr
            [description] => description_fr
        )

)

This would then allow you to insert individual entries for each language, by iterating over the $_POST form values, as an array of language associated form fields.

$sql = 'INSERT INTO `table_name` (`Name`, `Description`, `Language`) VALUES (?, ?, ?)';
if ($stmt = mysqli_prepare($conn, $sql)) {
    mysqli_stmt_bind_param($stmt, 'sss', $name, $description, $language);
    foreach ($_POST as $language => $fields) {
       $name = array_key_exists('name', $fields) ? $fields['name'] : null;
       $description = array_key_exists('description', $fields) ? $fields['description'] : null;
       mysqli_stmt_execute($stmt);
    }
}

---

Disclaimer: As there was little information provided on what the parent table relates to or its association(s), it was excluded from my answer. I presume the OP will be able to determine how to manipulate the answer to satisfy adding the parent to the query.