checkboxe og db

<?php
$q = "SELECT * FROM employee";
$r = mysql_query($q);

// show all employees and add checkboxes
while($row = @mysql_fetch_array($r)){
    echo '<input type="checkbox" name="employes[]" value="'.$row['Username'].'">'.$row["Username"].'<br />';
}

if ($_GET["op"] == "submit") {
    foreach ($_GET['employes'] AS $username) {
        echo $username.' checkbox afkrydset<br />';
    }
}
?>

Hvordan afkrydset?
Har du et felt i DB for det?

schaefner>> Tak for hjælpen, men det virker ikke helt.. for følgende:
Warning: Invalid argument supplied for foreach() in C:\XAMPP\xampp\htdocs\test\addgroupmembers.php on line 36

prøv at vise hele din form.

sukos>nope.Menigen er at man skal kunne afkrydes de employees som er hivet op fra datbasen, og hvis de er afkrydset skal de indsættes i en anden tabel i databasen

Ellers prøv:

if ($_GET["op"] == "submit") {
    if ($_POST['employes']) {
        foreach ($_POST['employes'] AS $username) {
            echo $username.' checkbox afkrydset<br />';
        }
    } else {
        echo 'Ingen valgt!';
    }
}

<?php
session_start();
include "dbConfig.php";

//setup query and run
$q = "SELECT * FROM employee";
$r = mysql_query($q);

// show all employees and add checkboxes
    while($row = @mysql_fetch_array($r))
  {
  echo "<input type=\"checkbox\" name=\"".$row["Username"]."\">".$row["Username"]."<br>";
  }

  if ($_GET["op"] == "submit")
  {
      //Det er her magien skal ske :)
  }

     

?>
<FORM NAME = "test" method ="POST" action="?op=submit">
<INPUT TYPE = "hidden" name="resultsPlease" value = "Right now">
<INPUT TYPE = "submit" name="submit" value="Submit">
</FORM>

Så brug:

foreach ($_POST['employes'] AS $username) {
  echo $username.' checkbox afkrydset<br />';
}

for stadig

Warning: Invalid argument supplied for foreach() in C:\XAMPP\xampp\htdocs\test\addgroupmembers.php on line 17

<?php
session_start();
include "dbConfig.php";

//setup query and run
$q = "SELECT * FROM employee";
$r = mysql_query($q);
?>
<FORM NAME = "test" method ="POST" action="?op=submit">
<INPUT TYPE = "hidden" name="resultsPlease" value = "Right now">
<?php
// show all employees and add checkboxes
    while($row = @mysql_fetch_array($r))
  {
  echo "<input type=\"checkbox\" name=\"employes[]\" value=\"".$row["Username"]."\">".$row["Username"]."<br>";
  }

  if (isset($_GET["op"]) && $_GET["op"] == "submit")
  {
      foreach ($_POST['employes'] AS $username) {
  echo $username.' checkbox afkrydset<br />';
}
  }

   

?>

<INPUT TYPE = "submit" name="submit" value="Submit">
</FORM>

Hov det skal jo være PODT og ikke GET!

Nåh, POST----------------------

<?php
session_start();
include("dbConfig.php");

//setup query and run
$q = "SELECT * FROM employee";
$r = mysql_query($q);

// show all employees and add checkboxes
while($row = @mysql_fetch_array($r)){
    echo '<input type="checkbox" name="employes[]" value="'.$row['Username'].'">'.$row["Username"].'<br />';
}

if (isset($_POST['send']){
    foreach ($_POST['employes'] AS $username) {
        echo $username.' checkbox afkrydset<br />';
    }
}
?>

<form name="test" method="post" action="">
<input type="hidden" name="send" value="1">
<INPUT TYPE = "hidden" name="resultsPlease" value = "Right now">
<INPUT TYPE = "submit" name="submit" value="Submit">
</FORM>

får parse error på linie 14 if(isset....

forstår ikke hvad $_POST['employes'] gør / er ?

<?php
session_start();
include("dbConfig.php");

//setup query and run
$q = "SELECT * FROM employee";
$r = mysql_query($q);

// show all employees and add checkboxes
while($row = @mysql_fetch_array($r)){
    echo '<input type="checkbox" name="employes[]" value="'.$row['Username'].'">'.$row["Username"].'<br />';
}

if (isset($_POST['send'])) {
    foreach ($_POST['employes'] AS $username) {
        echo $username.' checkbox afkrydset<br />';
    }
}
?>

<form name="test" method="post" action="">
<input type="hidden" name="send" value="1">
<INPUT TYPE = "hidden" name="resultsPlease" value = "Right now">
<INPUT TYPE = "submit" name="submit" value="Submit">
</FORM>

hmm, hvordan vil hente en checkbox som ligger uden for form-tags?
Kan man det? :O)

Du har fat i noget ;)

<form name="test" method="post" action="">
<input type="hidden" name="send" value="1">

<?php
session_start();
include("dbConfig.php");

//setup query and run
$q = "SELECT * FROM employee";
$r = mysql_query($q);

// show all employees and add checkboxes
while($row = @mysql_fetch_array($r)){
    echo '<input type="checkbox" name="employes[]" value="'.$row['Username'].'">'.$row["Username"].'<br />';
}
?>

<INPUT TYPE="hidden" name="resultsPlease" value="Right now">
<INPUT TYPE="submit" name="submit" value="Submit">
</FORM>

<?php
if (isset($_POST['send'])) {
    foreach ($_POST['employes'] AS $username) {
        echo $username.' checkbox afkrydset<br />';
    }
}
?>

<?php
echo"<form action=\"". $_SERVER["PHP_SELF"] ."\" method=\"post\">\n";

for($i = 1; $i <= 5; $i++) {
echo"<input type=\"checkbox\" name=\"navn[]\" value=\"". $i ."\"> var ". $i ."<br>\n";
}

echo"<input type=\"submit\">\n";
echo"</form>";

if(isset($_POST["navn"])) {
echo"<p>\n";
foreach($_POST["navn"] as $vaerdi) {
echo $vaerdi ."<br>\n";
    }
}
?>

<?php
session_start();
include("dbConfig.php");
?>

<form name="test" method="post" action="">
<input type="hidden" name="send" value="1">

<?php
//setup query and run
$q = "SELECT * FROM employee";
$r = mysql_query($q);

// show all employees and add checkboxes
while($row = @mysql_fetch_array($r)){
    echo '<input type="checkbox" name="employes[]" value="'.$row['Username'].'">'.$row["Username"].'<br />';
}
?>

<INPUT TYPE="hidden" name="resultsPlease" value="Right now">
<INPUT TYPE="submit" name="submit" value="Submit">
</FORM>

<?php
if (isset($_POST['send'])) {
    foreach ($_POST['employes'] AS $username) {
        echo $username.' checkbox afkrydset<br />';
    }
}
?>

schaefner>>helt super. Kan jeg istedet for at echo username, echo id'et i stedet for usernamet for employeen ?

<?php
session_start();
include("dbConfig.php");
?>

<form name="test" method="post" action="">
<input type="hidden" name="send" value="1">

<?php
//setup query and run
$q = "SELECT * FROM employee";
$r = mysql_query($q);

// show all employees and add checkboxes
while($row = @mysql_fetch_array($r)){
    echo '<input type="checkbox" name="employes[]" value="'.$row['id'].'">'.$row["Username"].'<br />';
}
?>

<INPUT TYPE="hidden" name="resultsPlease" value="Right now">
<INPUT TYPE="submit" name="submit" value="Submit">
</FORM>

<?php
if (isset($_POST['send'])) {
    foreach ($_POST['employes'] AS $id) {
        echo $id.' checkbox afkrydset<br />';
    }
}
?>

schaefner >> virkede ikke helt

hvad hedder dit felt med id'et?

I databasen hedder feltet Id

<?php
session_start();
include("dbConfig.php");
?>

<form name="test" method="post" action="">
<input type="hidden" name="send" value="1">

<?php
//setup query and run
$q = "SELECT * FROM employee";
$r = mysql_query($q);

// show all employees and add checkboxes
while($row = @mysql_fetch_array($r)){
    echo '<input type="checkbox" name="employes[]" value="'.$row['Id'].'">'.$row["Username"].'<br />';
}
?>

<INPUT TYPE="hidden" name="resultsPlease" value="Right now">
<INPUT TYPE="submit" name="submit" value="Submit">
</FORM>

<?php
if (isset($_POST['send'])) {
    foreach ($_POST['employes'] AS $id) {
        echo $id.' checkbox afkrydset<br />';
    }
}
?>

Helt genialt - en alle sidste lille bitte ting. Når jeg kommer ind på siden, står der checkbox afkrydset 2 gange, og når jeg reloader siden står der f. eks 1 og 3 afkrydset svarende det til er stod før jeg havde reloadet siden

den fik jeg ikke helt fat i.

når jeg reloader siden, står der hvad der blev afkrydset før reloaden. F.eks hvis jeg har afkrydet 1 og 3 og trykker submit, står der 1 og 3 afkrydset. Hvis jeg så reloader siden, står der stadig 1  og 3 afkrydset, i stedet for ingenting

Kommer der ikke en boks frem når du vil opdatere?

tror det er ligemeget. TUSIND tak for hjælpen.. her er 200 MEGET velfortjente point