当我有复选框中的数组时，MySQL foreach INSERT。

Question

I have a list of checkboxes that I want to insert into a database, which I have processed as an array:

<input name="my_array[]" value="<?php echo $data['ID'] ?>" type="checkbox">
<?php echo $data['name'] ?><?php echo "br/>";

I'm getting my list of checkboxes from my database, and the selected checkboxes are being posted like so:

   $strFoo=$_POST['my_array'];

    $strBar = "";


    foreach ($strFoo as $strFooName) {

    $strBar .= $strFooName . ", ";
    }
    $strBar = substr($strBar, 0, -2);

Echoing, I get my list of ID's selected like so 1, 2, 3, 4

My below foreach is unfortunately only inserting the first ID of the array only..

foreach ($strFoo as $strFooName) {

        $strSql="INSERT INTO table (ID) 
                 VALUES ('$strBar')";
    }

How can I insert each ID into my table?

Answer 1

It'd have to be:

INSERT INTO table (ID) VALUES (1), (2), (3), (4)

so your processing step would need to be

$values = array();
foreach ($strFoo as $strFooName) {
     $values[] = '(' . intval($strFooName) . ')';
}

$strBar = implode(',', $values);

Note the addition of intval - that ensures that only valid numbers get inserted into the SQL statement. Your version was vulnerable to SQL injection attacks.

Answer 2

$strSql="INSERT INTO table (ID) VALUES ('$strBar')";

应该

$strSql="INSERT INTO table (ID) VALUES ('$strFooName')";

Answer 3

您使用$strFooName在数组中循环，但您插入的是$strBar ，上面刚刚更新了一次。.也许是问题所在

Answer 4

From 12.2.5. INSERT Syntax

INSERT statements that use VALUES syntax can insert multiple rows. To do this, include multiple lists of column values, each enclosed within parentheses and separated by commas. Example:

INSERT INTO tbl_name (a,b,c) VALUES(1,2,3),(4,5,6),(7,8,9);

That said, however. It looks like you're trying to just loop straight through your post array with this piece of code

foreach ($strFoo as $strFooName) {
        $strSql="INSERT INTO table (ID) VALUES ('$strBar')";
}

$strBar in that example hasn't been created. You might try

foreach ($strFoo as $strFooName) {
        $strSql="INSERT INTO table (ID) VALUES ('$strFooName')";
}

Although you should also look into preventing SQL injection if you're going to do that.