What I'm trying to do is to have one datatable probably as a template and passing different datas on different pages.
Example functions.php
function force() {
global $DBconn;
$q = "SELECT * FROM table";
$res = $DBconn->query($q);
return $this->result;
}
function force_1() {
global $DBconn;
$q = "SELECT * FROM table_1";
$res = $DBconn->query($q);
return $this->result;
}
function table() {
echo '
<table class="sortable">
<thead>
<tr>
<th>
Title
</th>
</tr>
</thead>
<tbody>';
foreach ($this->result as $key) {
echo ' <tr>
<td>
$key->name
</td>
</tr>';
}
echo '</tbody>
</table>';
}
return true;
}
Then in the page.php
$table = table();
$smarty->assign("table", $table);
and page.tpl
<div> {$table} </div>
Currently nothing showed on page, not even an error or something. So:
Is this the way that should be done? With using the returned result in either function and then pass it to the smarty template?
I don't know how to use the smarty template (I never used it), but the $table variable holds what the function table() returns (in your case, the boolean value true). It should return a string, like this:
function force() {
// corrected the function force return value, assuming that you are using PDO connection
global $DBconn;
$q = "SELECT * FROM table";
$res = $DBconn->query($q);
return $res->fetchAll(PDO::FETCH_ASSOC);
}
function table() {
$arr = force();
return
'
<table class="sortable">
<thead>
<tr>
<th>
Title
</th>
</tr>
</thead>
<tbody>';
foreach ($arr as $row) {
echo " <tr>
<td>
$row['name']
</td>
</tr>";
}
echo '</tbody>
</table>';
}
You should be also careful with quotes when using variables inside a string.
EDIT:
If you are using mysqli class
function force() {
global $DBconn;
$q = "SELECT * FROM table";
return $DBconn->query($q);
}
function table() {
$result = force();
return
'
<table class="sortable">
<thead>
<tr>
<th>
Title
</th>
</tr>
</thead>
<tbody>';
//while ($row = $result->fetch_assoc()) {
while ($row = mysql_fetch_assoc($result)) {
echo " <tr>
<td>
$row['name']
</td>
</tr>";
}
echo '</tbody>
</table>';
}
I am using mysql queries to fetch data from db. All my data are showing fine in tables. Now I want to color the status by value less than 2 or more than 3.
The below code is not working.
Need help.
$result = mysqli_query($connect, $query);
if(mysqli_num_rows($result) > 0)
{
$output .= '<div class="table-responsive">
<table class="table table bordered">
<tr>
<th>name</th>
<th>genre</th>
<th>time</th>
<th>status</th>
<th>more...</th>
</tr>';
?>
<?php
function status_style($row) {
if ($row < 2) return 'background-color: #ff0000'; //red
if ($row > 3) return 'background-color: #33cc33'; //green
return '';
}
?>
<?php
while($row = mysqli_fetch_array($result))
{
$output .= '
<tr>
<td>'.$row["name"].'</td>
<td>'.$row["genre"].'</td>
<td>'.$row["time"].'</td>
<td>'.$row["status"].'</td>
<td>'.$row["more"].'</td>
</tr>
';
}
echo $output;
{
echo 'Data Not Found';
}
?>
You have already created function to color the rows but haven't called the function during showing output. That's why it is not working.
Change your $output variable with:
$output .= '
<tr>
<td>'.$row["name"].'</td>
<td>'.$row["genre"].'</td>
<td>'.$row["time"].'</td>
<td style="'.status_style($row["status"]).'">'.$row["status"].'</td>
<td>'.$row["more"].'</td>
</tr>
';
You can try something like this:
<?php
$result = mysqli_query($connect, $query);
if(mysqli_num_rows($result) > 0)
{
$output .= '<div class="table-responsive">
<table class="table table bordered">
<tr>
<th>name</th>
<th>genre</th>
<th>time</th>
<th>status</th>
<th>more...</th>
</tr>';
function status_style($row) {
if ($row < 2) return $color = '#ff0000'; //red
if ($row > 3) return $color = '#33cc33'; //green
return $color = '';
}
while($row = mysqli_fetch_array($result))
{
$output .= '
<tr style='"'background-color: '"' . $color . '"'>
<td>'.$row["name"].'</td>
<td>'.$row["genre"].'</td>
<td>'.$row["time"].'</td>
<td>'.$row["status"].'</td>
<td>'.$row["more"].'</td>
</tr>
';
}
echo $output;
{
echo 'Data Not Found';
}
You didn't do anything with the background color.
HTML Table getting mixed up when having echo in while loop. Bellow attached the output image for that. How can i get normal table with loop result? Please note i am getting table data by searching with manual date.
<?php
if (isset($_POST['submit_date'])) {
if (empty($_POST['m_date'])) {
echo '<div class="alert alert-danger">Error: Select date then search</div>';
}else{
$m_date = $_POST['m_date'];
$q = mysqli_query($conn, "SELECT * FROM bazar_dor WHERE m_date='$m_date'");
echo '<table style="width:100%">
<tr>
<th>Category Name</th>
<th>Price</th>
</tr>
<tr>
';
while ($row=mysqli_fetch_array($q)) {
$cat_name = $row['cat_name'];
$price = $row['price'];
echo '<td>'.$cat_name.'</td><td>'.$price.'</td>';
}
echo '</tr></table>';
}
}
?>
the output i am getting from this code
It looks like you'll need to include table rows (<tr>) inside your loop, as well. Your current code outputs all data on the same row, which undoubtedly breaks the table.
echo '<table style="width:100%">
<tr>
<th>Category Name</th>
<th>Price</th>
</tr>';
while ($row=mysqli_fetch_array($q)) {
$cat_name = $row['cat_name'];
$price = $row['price'];
echo '<tr><td>'.$cat_name.'</td><td>'.$price.'</td></tr>';
}
echo '</table>';
Replace your cede with this
<?php if (isset($_POST['submit_date'])) { if (empty($_POST['m_date'])) { echo '<div
class="alert alert-danger">Error: Select date then search</div>'; }else{ $m_date = $_POST['m_date'];
$q = mysqli_query($conn, "SELECT * FROM bazar_dor WHERE m_date='$m_date'");
echo '<table style="width:100%"> <tr>
<th>Category Name</th> <th>Price</th> </tr> ';
while ($row=mysqli_fetch_array($q)) { $cat_name = $row['cat_name'];
$price = $row['price']; echo '<tr><td>'.$cat_name.'</td><td>'.$price.'</td><\tr>';
} echo '</table>';
} } ?>
I am coding php and seen this error while I trying to do a HTM5 validator. But the things is there is no in my whole file. Any suggestions what is wrong here?
Error Line 27, Column 4: Start tag em seen in table.
<em><br />
Edit note : I rarely want to add this code here due to the fact that's it is long and have nothing to do with the error (like I said above, there is no tag in the html)
<section>
<table>
<?php
include("./settings.php");
$conn = #mysqli_connect($host, $user, $pwd, $sql_db);
if (!$conn) {
echo "<p>Databse connection error</p>";
} else {
$eoinumber = trim($_POST["num"]);
$lname = trim($_POST["lname"]);
$query = "SELECT * FROM eoi WHERE EOInumber = $eoinumber AND lname = '$lname'";
$result = mysqli_query($conn,$query);
if (!$result) {
echo "<p>There is something wrong with the $query</p>";
} else {
$temp = mysqli_num_rows($result);
if ($temp == 0) {?>
<p>0 enquiry found. Please check again your EOInumber and last name</p>
<?php } else {
?>
<thead>
<tr>
<th>EOI number</th>
<th>Job ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Date of Birth</th>
<th>Gender</th>
<th>Street</th>
<th>Town</th>
<th>State</th>
<th>Postcode</th>
<th>Mail</th>
<th>Telephone</th>
<th>Skills</th>
<th>Other Skills</th>
</tr>
</thead>
<?php
while ($row = mysqli_fetch_assoc($result)) {
echo "<tbody>";
echo "<tr>";
echo "<td>",$row["EOInumber"],"</td>";
echo "<td>",$row["job_num"],"</td>";
echo "<td>",$row["fname"],"</td>";
echo "<td>",$row["lname"],"</td>";
echo "<td>",$row["bday"],"</td>";
echo "<td>",$row["gender"],"</td>";
echo "<td>",$row["street"],"</td>";
echo "<td>",$row["town"],"</td>";
echo "<td>",$row["state"],"</td>";
echo "<td>",$row["postcode"],"</td>";
echo "<td>",$row["mail"],"</td>";
echo "<td>",$row["tele"],"</td>";
echo "<td>",$row["skill"],"</td>";
echo "<td>",$row["other"],"</td>";
echo "</tr>";
echo "</tbody>";
// echo "<p>Sucess</p>";
}
?>
</table>
Regardless of the presence or otherwise of the <em> tag, you are going to run into problems because you are opening the table tag and then potentially putting a <p> or other elements directly inside it, which is illegal (you can put <thead>, <tbody>, <caption>, <tr>, <tfoot> elements inside a <table>). It would be better if you opened the table after checking whether the queries had run successfully or not, i.e.:
<section>
<?php
include("./settings.php");
$conn = #mysqli_connect($host, $user, $pwd, $sql_db);
if (!$conn) {
echo "<p>Databse connection error</p>";
} else {
$eoinumber = trim($_POST["num"]);
$lname = trim($_POST["lname"]);
$query = "SELECT * FROM eoi WHERE EOInumber = $eoinumber AND lname = '$lname'";
$result = mysqli_query($conn,$query);
if (!$result) {
echo "<p>There is something wrong with the $query</p>";
} else {
$temp = mysqli_num_rows($result);
if ($temp == 0) {
echo "<p>0 enquiry found. Please check again your EOInumber and last name</p>";
} else {
echo "<table>";
... echo the table header...
while ($row = mysqli_fetch_assoc($result)) {
... echo table results...
}
echo "</table>";
}
}
?>
You missed a few }
So this should work:
<section>
<table>
<?php
include("./settings.php");
$conn = #mysqli_connect($host, $user, $pwd, $sql_db);
if (!$conn) {
echo "<p>Databse connection error</p>";
exit;
}
$eoinumber = trim($_POST["num"]);
$lname = trim($_POST["lname"]);
$query = "SELECT * FROM eoi WHERE EOInumber = $eoinumber AND lname = '$lname'";
$result = mysqli_query($conn,$query);
if (!$result) {
echo "<p>There is something wrong with the $query</p>";
exit;
} else {
$temp = mysqli_num_rows($result);
}
if ($temp == 0) { ?>
<p>0 enquiry found. Please check again your EOInumber and last name</p>
<?php } else { ?>
<thead>
<tr>
<th>EOI number</th>
<th>Job ID</th>
<th>First Name</th>
<th>Last Name</th>
<th>Date of Birth</th>
<th>Gender</th>
<th>Street</th>
<th>Town</th>
<th>State</th>
<th>Postcode</th>
<th>Mail</th>
<th>Telephone</th>
<th>Skills</th>
<th>Other Skills</th>
</tr>
</thead>
<?php
while ($row = mysqli_fetch_assoc($result)) {
echo "<tbody>";
echo "<tr>";
echo "<td>",$row["EOInumber"],"</td>";
echo "<td>",$row["job_num"],"</td>";
echo "<td>",$row["fname"],"</td>";
echo "<td>",$row["lname"],"</td>";
echo "<td>",$row["bday"],"</td>";
echo "<td>",$row["gender"],"</td>";
echo "<td>",$row["street"],"</td>";
echo "<td>",$row["town"],"</td>";
echo "<td>",$row["state"],"</td>";
echo "<td>",$row["postcode"],"</td>";
echo "<td>",$row["mail"],"</td>";
echo "<td>",$row["tele"],"</td>";
echo "<td>",$row["skill"],"</td>";
echo "<td>",$row["other"],"</td>";
echo "</tr>";
echo "</tbody>";
// echo "<p>Sucess</p>";
}
}
?>
</table>
<!--...-->
Im writing a function to get the parent and child ids but for the third loop there is a
problem the loop gets even the previous loops id also .
How can i avoid it?
<?
$results = '
<table>
<thead>
<tr >
<td id="ticket" align="center" ><b>Task<br />ID</b></td>
<td id="ticket" align="center" ><b>col1</td>
<td id="ticket" align="center" ><b>col2</td>
</tr>
</thead>
<tbody>';
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC))
{
$results .='
<tr >
<td align="center">
'.$row['Task_id'].'
</td>';
$results .= '<td align="center">';
$gg = mysqli_query($dbc,"select * from Tasks where ParentTask_Id='".$row['Task_id']."'");
echo "<br>";
while ($rowdd = mysqli_fetch_assoc($gg))
{
$results .= $rowdd['Task_id']."<br><br>";
$gg2 = mysqli_query($dbc,"select * from Tasks where ParentTask_Id='".$rowdd['Task_id']."'");
while ($rowdd2 = mysqli_fetch_assoc($gg2))
{
$results2 = $rowdd2['Task_id']."<br><br>";
}
echo "<br>";
}
// $results .= $car ;
// $results .= $t;
$results .='</td>';
$results .=' <td align="left" >'?>
<?
$results .= $results2;
$results .='</td>';
$results .='
</tr>';
}
?>
Is the $results variable empty? I only see it being concatenated.
Also, on your table you have multiple ids that are the same. You either need to change that to a class or have a unique value for each id.