我试图通过PHP代码更新记录。 但我的代码有问题

Question

I am trying to update records by php code. but having trouble in my code.

code is running without using functions (get40(),show()) but i want to use buttons to call these functions. code is give below:

this is main code file:

    <?php
$data;

    $count=0;
    $host='localhost';
    $uname='root';
    $pass="";
    $dbname="testing";
    $db=new mysqli($host,$uname,$pass,$dbname);
    $result;
    $query;


    if (mysqli_connect_errno()) {
    echo 'Error: Could not connect to database. Please try again later.';
    exit;
    }

    function get40($off='0', $rcd='2')
    {
        global $db,$result,$query;
        $query="SELECT sword, tword, reviewed from TempDictionary LIMIT $off,$rcd";
        $result = $db->query($query)or trigger_error($db->error);
    }

    function show()
    {
        global $db,$result,$query;
        echo "<form action='testSave.php' method='POST'>";
        echo "<table border=1 >";
        while($row = mysqli_fetch_array($result)) 
        {
            global $count;
            $count+=1;
            echo "<tr>";
            echo "<td><input type='text' name='s[]' value=" . $row['sword'] . "></td>";
            echo "<td><input type='text' name='t[]' value=" . $row['tword'] . "></td>";
            echo "<td><input type='text' name='r[]' size='1' value=" .$row['reviewed'] . "></td>";
            echo "</tr>";
        }
        echo "</table>";
        echo "<input type='text' name='no' size='2' value=" .$count. ">";
        echo "<input type='submit' value='Save'>";
        echo "</form>";
}
//get40(1,2);
//show();
mysqli_close($db);
?>
<script type="text/javascript"> 
function get()
{
 <?php 
    get40(0,3);
    show();
 ?>
 }
</script>

<form action="" method="POST">
<input type=button name=btnGet value=Get onclick="get()">
</form>

and savetest.php file is:

    <?php

$no=$_POST['no'];
$i=0;
$s=$_POST['s']; 
$t=$_POST['t'];
$r=$_POST['r'];

    $host='localhost';
    $uname='root';
    $pass="";
    $dbname="testing";
    $db=new mysqli($host,$uname,$pass,$dbname);

    if (mysqli_connect_errno()) {
    echo 'Error: Could not connect to database. Please try again later.';
    exit;
    }

    // display
    foreach ($s as $key=>$value) 
    {
        print "$key = $value $t[$key] $r[$key]";
        echo '<br>';
    }

    // save to DB
    foreach ($s as $key=>$value) 
    {
    //  $query="UPDATE TempDictionary set sword=".$value.",tword=".$t[$key].",reviewed=".$r[$key]."
    //          WHERE sword=".$value;
        $query="UPDATE TempDictionary set sword='$value',tword='$t[$key]',reviewed='$r[$key]'
                WHERE sword='$value'";              

    $result = $db->query($query)or trigger_error($db->error);

    }

mysqli_close($db);  
?>

the problem is while i click button it does not works.

Answer 1

AFAIK you are trying to run PHP code with Javascript code. Use Jquery/Ajax to solve your problem. You can Google for that or look at stackoverflow.

Step 1: Learn how a click on a button can run PHP Run php function inside jQuery click

Step 2: Learn how to use data from a database and populate a form with it: ajax call to populate form fields from database query when select value changes

That's the first step in solving your problem. (sorry it's no copy/paste solution, just a pointer in the right direction)