CODEWITHSUNDEEP
phpjquery
DisplayName
DisplayName

Reputation: 51

HTML table from output

This may be very basic but I have searched for over a day trying various different things, all to no avail.

I am trying to output more than 1 row if more than 1 row exists. So far it is only showing me the last row!

My code is as follows:

$query = "SELECT CaseID FROM Cases WHERE ClientID='".$_GET['ClientID']."'";
$result = mysqli_query($con, $query);
while($row = mysqli_fetch_array($result))
{
    $CaseID = $row['CaseID'];
}

$result = mysqli_query($con,"SELECT dbUser FROM SalesUser WHERE ID='".$SalesID."';");
while($row = mysqli_fetch_array($result))
{
    $SalesUser = $row['dbUser'];
}

?>

<table class="table table-striped">
  <thead>
    <tr>
      <th>Case ID</th>
      <th>Sales User</th>
    </tr>
  </thead>
  <tbody>
    <tr>
      <td><? echo $CaseID; ?></td>
      <td><? echo $SalesUser; ?></td>
    </tr>
  </tbody>
</table>

Basically, I want to search for a Client ID, and have it return all cases which are for that particular client.

Upvotes: 0

Views: 58

Answers (2)

DisplayName
DisplayName

Reputation: 51

For anyone else who may run into this, @roberto06 kindly assisted and it has now been fixed and is working.

$query = "SELECT CaseID, Provider, FROM Cases WHERE ClientID='".$_GET['ClientID']."'";
$result = mysqli_query($con, $query);
$resultsArray = array();
while($row = mysqli_fetch_array($result))
{
    $resultArray = array('CaseID' => $row['CaseID'], 'Provider' => $row['Provider'] );
    $resultsArray[] = $resultArray;
}

?>

<table class="table table-striped">
    <thead>
        <tr>
            <th>Case ID</th>
            <th>Provider</th>
        </tr>
    </thead>
    <tbody>
    <?php foreach ($resultsArray as $resultArray) { ?>
        <tr>
            <td><? echo $resultArray['CaseID']; ?></td>
            <td><? echo $resultArray['Provider']; ?></td>
        </tr>
    <?php } ?>
    </tbody>
</table>

Upvotes: 0

roberto06
roberto06

Reputation: 3864

You have to store your caseID rows into an array and then loop over it : 

<?php

$query = "SELECT CaseID FROM Cases WHERE ClientID='".$_GET['ClientID']."'";
$result = mysqli_query($con, $query);
$CaseIDs = array();
while($row = mysqli_fetch_array($result)) {
    $CaseIDs[] = $row['CaseID'];
}

$result = mysqli_query($con,"SELECT dbUser FROM SalesUser WHERE ID='".$SalesID."';");
while($row = mysqli_fetch_array($result)) {
    $SalesUser = $row['dbUser'];
}

?>

<table class="table table-striped">
    <thead>
        <tr>
            <th>Case ID</th>
        </tr>
    </thead>
    <tbody>
    <?php foreach ($CaseIDs as $CaseID) { ?>
        <tr>
            <td><? echo $CaseID; ?></td>
        </tr>
    <?php } ?>
    </tbody>
</table>

EDIT : Same goes for SalesID.

EDIT 2 : With only one array : 

<?php

$query = "SELECT Cases.CaseID, SalesUser.dbUser FROM Cases LEFT JOIN SalesUser ON [...] WHERE [...] "; // Can't help you with your query as I don't know your DB structure
$result = mysqli_query($con, $query);
$resultsArray = array();
while($row = mysqli_fetch_array($result)) {
    $resultArray = array('CaseID' => $row['CaseID'], 'dbUser' => $row['dbUser']);
    $resultsArray[] = $resultArray;
}

?>

<table class="table table-striped">
    <thead>
        <tr>
            <th>Case ID</th>
            <th>Sales User</th>
        </tr>
    </thead>
    <tbody>
    <?php foreach ($resultsArray as $resultArray) { ?>
        <tr>
            <td><?php echo $resultArray['CaseID']; ?></td>
            <td><?php echo $resultArray['dbUser']; ?></td>
        </tr>
    <?php } ?>
    </tbody>
</table>

Upvotes: 1

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