CODEWITHSUNDEEP
phploopsforeach
DiegoP.
DiegoP.

Reputation: 45737

PHP Foreach loop problem with mySQL INSERT INTO

I am having a big issue. This is the first time I sue a foreach and I do not even know if it's the right thing to use. I have a textarea where my members can add some text. They also have all the accounts where to send the posted text. Accounts are of two types F and T. They are shown as checkboxes. So when a member types "submit" the text should be INSERTED in a specific table for EACH of the selected accounts. I thought php foreach was the right thing. But I am not sure anymore. Please take in mind I do not know anything about foreach and arrays. So please when helping me, consider to provide the modded code =D . Thank you so much!

<?php
require_once('dbconnection.php');
$MembID = (int)$_COOKIE['Loggedin'];
?>
<form action="" method="post">
    <p align="center"><textarea id="countable1" name="addit" cols="48" rows="10" style="border-color: #ccc; border-style: solid;"></textarea>
    <br>
<?
$DB = new DBConfig();
$DB -> config();
$DB -> conn();
$on="on";
    $queryF ="SELECT * FROM `TableF` WHERE `Active`='$on' AND `memberID`='$MembID' ORDER BY ID ASC";
    $result=mysql_query($queryF) or die("Errore select from TableF: ".mysql_error());
$count = mysql_num_rows($result); 
if ($count > 0)
    {
    while($row = mysql_fetch_array($result)) {
    ?><div style="width:400px; height:100px;margin-bottom:50px;"><?
    $rowid = $row['ID'];
    echo $row['Name'] . '</br>';
    $checkit = "checked";
    if ($row['Main'] == "")
    $checkit = "";
    if ($row['Locale'] =="")
    $row['Locale'] ="None" . '</br>';
    echo $row['Locale'] . '</br>';
    if ($row['Link'] =="")
    $row['Link'] ="javaScript:void(0);";
    ?>
    <!-- HERE WE HAVE THE "F" CHECKBOXES. $rowid SHOULD BE TAKEN AND INSERTED IN THE FOREACH BELOW FOR ANY SELECTED CHECKBOX IN THE FIELD "Type".SEE BELOW -->
    <input type="checkbox" name="f[<?php echo $rowid?>]" <?php echo $checkit;?>>
    </div>
    <?
}//END WHILE MYSQL
}else{
echo "you do not have any Active account";
}
$DB = new DBConfig();
$DB -> config();
$DB -> conn();
    $queryTW ="SELECT * FROM `TableT` WHERE `Active`='$on' AND `memberID`='$MembID' ORDER BY ID ASC";
    $result=mysql_query($queryTW) or die("Errore select TableT: ".mysql_error());
$count = mysql_num_rows($result); 
if ($count > 0)
    {
    while($row = mysql_fetch_array($result)) {
    ?><div style="width:400px; height:100px;margin-bottom:50px;"><?
    $rowid = $row['ID'];
    echo $row['Name'] . '</br>';
    $checkit = "checked";
    if ($row['Main'] == "")
    $checkit = "";
    if ($row['Locale'] =="")
    $row['Locale'] ="None" . '</br>';
    echo $row['Locale'] . '</br>';
    if ($row['Link'] =="")
    $row['Link'] ="javaScript:void(0);";
    ?>
        <!-- HERE WE HAVE THE "T" CHECKBOXES. $rowid SHOULD BE TAKEN AND INSERTED IN THE FOREACH BELOW FOR ANY SELECTED CHECKBOX IN THE FIELD "Type".SEE BELOW -->
    <input type="checkbox"  name="t[<?php echo $rowid?>]" <?php echo $checkit;?> >
    </div>
    <?
}//END 2° WHILE MYSQL
}else{
echo "you do not have any Active account";
}
?>
<input type="submit" value="submit">
</form>
<?
//WHEN CHECKBOXES "F" ARE FOUND, FOR EACH CHECKBOX IT SHOULD INSERT INTO TableG THE VALUES BELOW, FOR EACH SELECTED "F" CHECKBOX
if(!empty($_POST['addit']) && !empty($_POST['f'])){
$thispostF = $_POST['f'];
$f="F";
foreach ($thispostF as $valF)
//THE MOST IMPORTANT FIELD HERE IS "Type". THE ARRAY INSERT "on", I NEED INSTEAD THE VALUE $rowid AS ABOVE
$queryaddF="INSERT INTO TableG (ID,memberID,Type,IDAccount,Tuitting) VALUES (NULL,".$MembID.",'".$f."','".$valF."', '".$_POST['addit']."')";
$resultaddF=mysql_query($queryaddF) or die("Errore insert G: ".mysql_error());
}
//WHEN CHECKBOXES "T" ARE FOUND, FOR EACH CHECKBOX IT SHOULD INSERT INTO TableG THE VALUES BELOW, FOR EACH SELECTED "T" CHECKBOX
if(!empty($_POST['addit']) && !empty($_POST['t'])){
$thispostT = $_POST['t'];
$t="T";
foreach ($thispostT as $valF)
//THE MOST IMPORTANT VALUE HERE IS "Type". THE ARRAY GIVES "on", I NEED INSTEAD THE VALUE $rowid AS ABOVE
$queryaddT="INSERT INTO TableG (ID,memberID,Type,IDAccount,Tuitting) VALUES (NULL,".$MembID.",'".$t."','".$valF."', '".$_POST['addit']."')";
$resultaddF=mysql_query($queryaddF) or die("Errore insert G: ".mysql_error());
}
?>

Upvotes: 0

Views: 5903

Answers (3)

Syed Shah
Syed Shah

Reputation: 1

My suggestion is

create field name as t[] (array) onchecked value will be passed on the next page

the form checkbox field should be like that

<input type="checkbox"  name="t[]" value="< ?php echo $rowid?>" <?php echo $checkit;? > >

and when you Submit the form 

GET THE VALUE and insert in to database;
< ?
if($_POST['t'])
{
foreach($_POST['t'] as $v)
{
queryaddT="INSERT INTO TableG (ID,memberID,Type,IDAccount,Tuitting) VALUES (NULL,".$MembID.",'".$t."','".$valF."', '".$_POST['addit']."')";
$resultaddF=mysql_query($queryaddF) or die("Errore insert G: ".mysql_error());


}

}
? >

Upvotes: 0

David Nguyen
David Nguyen

Reputation: 8528

I think I know what you're doing...

You're going to need to do:

foreach($_POST as $key => $value) {
   $type = substr($key,0,1);
   $id = substr($key,1);
   if($type == 't') {
      // do insert for t table here
   } else if($type == 'f') {
      // do insert for f table here
   }
}

I didn't test it but it should be something like this.

Upvotes: 1

Khurram Ijaz
Khurram Ijaz

Reputation: 1864

 foreach ($thispostT as $valF)
{

$queryaddT="INSERT INTO TableG (ID,memberID,Type,IDAccount,Tuitting) VALUES (NULL,".$MembID.",'".$t."','".$valF."', '".$_POST['addit']."')";
$resultaddF=mysql_query($queryaddF) or die("Errore insert G: ".mysql_error());
}

please put start and ending bracket to your foreach loop and try i have not read the whole code but just found you missing the brackets. hope that helps you.

Upvotes: 6

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