PHP insert the array values into mysql

Question

I had a problem inserting arrays into a MySQL database. I think found out that I cannot insert an array into columns of the table, and I should separate the values of the array and then do the insert action. But I don't know how to separate the values and insert those values. Should I separate the values and not use array?

Also I would like to create the table to shows all values(number1-10) of db stored.

Thanks for everyone!

Here's my codes below:

  $varNum = array('1','2','3','4','5','6','7','8','9','10');
  //an array showed in the selection box

  <form action="testing_Rand.php" method="post">
    <p><b><center>Choose an amount of random numbers in the selection box</center></b></p>
    <p>
            <select name="selectNum">
                <?php
                    foreach($varNum as $key => $value):
                    echo '<option value="'.$key.'">'.$value.'</option>';
                    endforeach;
                ?>
            </select>
    </p>
    <center><input type="Submit" value="submit"></center>
</form> 
    //A POST function to generate the random numbers


    //do post function
    if(isset($_POST["selectNum"]) ){
    $arrayRand=intval($_POST["selectNum"]);    
    for($i=0;$i<=$arrayRand;$i++){
         $varNum[$i]=rand(1,10000);    
    } 

    var_dump($varNum);  //show results  
    $newRand = "INSERT INTO testing_Rand (number1, number2, number3, number4, number5, number6, number7, number8, number9, number10) VALUES ('$varNum')";
    mysql_query($newRand);
}

    //show mysql database results

        $sqlDBrand = "SELECT id, number1, number2, number3, number4, number5, number6, number7, number8, number9, number10 FROM rand.testing";
$result = mysql_query($sqlDBrand) or die('MySQL query error');
while($row = mysql_fetch_array($result)){
    echo $row['id'];
    echo $row['number1'];
    echo $row['number2'];
    echo $row['number3'];
    echo $row['number4'];
    echo $row['number5'];
    echo $row['number6'];
    echo $row['number7'];
    echo $row['number8'];
    echo $row['number9'];
    echo $row['number10'];
}

Answer 1

i think this may help you, first make a string $sql and then use it in mysql_query()

  $sql="INSERT INTO rand.testing (number1, number2, number3, number4, number5, number6, number7, number8, number9, number10) VALUES (";
       $sql.=implode(',', $varNum);
       $sql.=")";
mysql_query($sql);

Answer 2

It is depending on your Data and your Goal.

Take a look at the function implode:

$arr = [1,2,3,4,5];
$data = implode ('|', [1,2,3,4,5]) //1|2|3|4|5
$arr = explode('|', $data); //Converted back

http://www.php.net/manual/de/function.implode.php

If you need it searchable, considering a serialization to XML

$arr = [1,2,3,4,5];
$xml = new SimpleXMLElement('<root/>');
array_walk_recursive($arr, array ($xml, 'addChild'));

https://stackoverflow.com/a/1397164/2441442

If you need it not searchable, you can also take a look at serialize

$arr = [1,2,3,4,5];
$data = serialize($arr);
unserialize($data);

http://php.net/manual/de/function.serialize.php

Answer 3

Going back a stage, you have a table with a fixed number of columns but a variable number of fields of data to store. This is generally a bad idea.

It would be better to split this into 2 tables. One just stores something to identify the group of numbers, and another table that has several rows for each group of numbers; one row per random number per group.

This is a basic part of normalisation of data.

You could insert the to the 2 tables and grab the values with something like this:-

<?php
    //A POST function to generate the random numbers


    //do post function
    if(isset($_POST["selectNum"]) )
    {
        $arrayRand=intval($_POST["selectNum"]);    
        for($i=0;$i<=$arrayRand;$i++)
        {
            $varNum[$i]=rand(1,10000);    
        } 

        if (count($varNum))
        {
            var_dump($varNum);  //show results  
            $newRand = "INSERT INTO testing_rand (id) VALUES(NULL)";
            mysql_query($newRand);
            $inserted_id = mysql_insert_id ();
            $newRand = "INSERT INTO testing_rand_numbers (rand_id, number) VALUES($inserted_id, ".implode("),($inserted_id, ", $varNum).")";
        }
    }

    //show mysql database results

    $sqlDBrand = "SELECT id, GROUP_CONCAT(b.number) AS all_numbers FROM testing_rand a LEFT OUTER JOIN testing_rand_numbers b ON a.id = b.rand_id";
    $result = mysql_query($sqlDBrand) or die('MySQL query error');
    while($row = mysql_fetch_array($result))
    {
        echo $row['id'];
        echo $row['all_numbers'];
    }
?>