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I am trying to create table in PHP using array but I am getting parse error in my foreach line. Don't know why its giving me syntax error.
$message = '<table border=1>
<tr>
<th><b>Order Details</b></th>
</tr>
'.foreach ($json_array->request->list as $value) {
$wineId = $value->wineId;
$wineName = $value->wineName;
$noOfDrinks = $value->noOfDrinks;
.'
<tr>
<td>ID</td><td>'.$wineId.'</td>
</tr>
<tr>
<td>Name</td><td>'.$wineName.'</td>
</tr>
<tr>
<td>Quantity</td><td>'.$noOfDrinks.'</td>
</tr>
'. } .'
</table>';
Try this, use string concatenation except php code
$message = '<table border=1>
<tr>
<th><b>Order Details</b></th>
</tr>';
foreach ($json_array->request->list as $value) {
$wineId = $value->wineId;
$wineName = $value->wineName;
$noOfDrinks = $value->noOfDrinks;
$message .= '<tr>
<td>ID</td><td>'.$wineId.'</td>
</tr>
<tr>
<td>Name</td><td>'.$wineName.'</td>
</tr>
<tr>
<td>Quantity</td><td>'.$noOfDrinks.'</td>
</tr>';
}
$message .='</table>';
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I'm trying to display products from cart but I receive following errors (see below). I really don't understand what I'm doing wrong... A detailed answer with explanations would be nice. I tried to find solutions to my problem without success. I do understand what an undefined variable means but I don't know how to fix it in this particular case.
Notice: Undefined variable: row
Warning: Invalid argument supplied for foreach()
<html>
<body>
<table>
<tbody>
<?php
require_once 'core/config.php';
$cartQ = $db->query("SELECT * FROM cart");
$result = mysqli_fetch_assoc($cartQ);
foreach ($row as $product) {
$productQuery = $db->query("SELECT * FROM product INNER JOIN cart ON product.id=cart.product_id ");
$product = mysqli_fetch_assoc($productQuery);
while ($row = mysqli_fetch_assoc($cartQ)) {
}
?>
<tr class="p">
<td class="image"><img src="<?= $product['image_1']; ?>" /></td>
<td class="name"><?= $product['prod_name']; ?></td>
<td class="price"><?= money($product['price']); ?></td>
</tr>
<?php
}
?>
</tbody>
</table>
</body>
</html>
Get rid of the first query and the foreach loop, they're not needed. The JOIN query returns all the information you need to display.
<html>
<body>
<table>
<tbody>
<?php
require_once 'core/config.php';
$productQuery = $db->query("SELECT * FROM product INNER JOIN cart ON product.id=cart.product_id ");
while ($product = mysqli_fetch_assoc($productQuery)) {
?>
<tr class="p">
<td class="image"><img src="<?= $product['image_1']; ?>" /></td>
<td class="name"><?= $product['prod_name']; ?></td>
<td class="price"><?= money($product['price']); ?></td>
</tr>
<?php
}
?>
</tbody>
</table>
</body>
</html>
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Closed 7 years ago.
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Can anyone help me how to display table inside PHP if statement.
Here's my code:
if( empty($errors))
{
ACTIVITIES \n
echo ' <table>
<thead>
<tr>
<th>Activity</th>
<th>Price</th>
<th>Quantity</th>
<th>Amount</th>
</tr>
</thead>
<tbody>
<tr>
<td>Chicken Feeding</td>
<td>$price_chicken</td>
<td>$num_chicken</td>
<td>$total_chicken</td>
</tr>
</tbody>
<tbody>
<tr>
<td>Fish Feeding</td>
<td>$price_fish</td>
<td>$num_fish</td>
<td>$total_fish</td>
</tr>
</tbody>
.................
</table> ';
}
try this
<?php if( empty($errors)): ?>
ACTIVITIES \n
<table>
<thead>
<tr>
<th>Activity</th>
<th>Price</th>
<th>Quantity</th>
<th>Amount</th>
</tr>
</thead>
<tbody>
<tr>
<td>Chicken Feeding</td>
<td><?=$price_chicken;?></td>
<td><?=$num_chicken;?></td>
<td><?=$total_chicken;?></td>
</tr>
</tbody>
<tbody>
<tr>
<td>Fish Feeding</td>
<td><?=$price_fish;?></td>
<td><?=$num_fish;?></td>
<td><?=$total_fish;?></td>
</tr>
</tbody>
.................
</table>
<?php endif; ?>
You have the table code wrapped in single quotes, but there are PHP variables inside. PHP will not translate those variables unless you wrap the table in double quotes.
If this doesn't help, give us an idea of what the output looks like.
Try using a 2-dimensional array and looping through it to create a proper HTML table. This way, there's no hard-coding, so you can make as many rows/columns as you want. Here's the code, just change the array:
<table>
<tbody>
<?php
$tableArray = [["Chicken Feeding", $price_chicken, $num_chicken],
["Fish Feeding", $price_fish, $num_fish]];
foreach ($tableArray as $tableRow) {
echo "<tr>";
foreach ($tableRow as $tableCell) {
echo "<td>$tableCell</td>";
}
echo "</tr>";
}
?>
</tbody>
</table>
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Closed 8 years ago.
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Can anybody point out what I am doing wrong in my code below? I'm getting an "UNEXPECTED T_WHILE" error.
echo '
<table border="1">
<tr>
<td>Events</td>
<td>Category</td>
</tr>
<tr>
<td>', while ($row = mysql_fetch_assoc($queryresult)) { $title = $row['eventTitle']; echo $title; }, '</td>
<td>', while ($row = mysql_fetch_assoc($queryresult)) { $category = $row['eventTitle']; echo $category; }, '</td>
</tr>
</table> ';
Try this
First off, close your PHP tags.
?>
<table border="1">
<tr>
<td>Events</td>
<td>Category</td>
</tr>
<tr>
<td>
<?php
while ($row = mysql_fetch_assoc($queryresult)) {
$title = $row['eventTitle'];
echo $title;
}
?>
</td>
<td>
<?php
while ($row = mysql_fetch_assoc($queryresult)) {
$category = $row['eventTitle'];
echo $category;
}
?>
</td>
</tr>
</table>
PHP -
$array = array();
$i = 0;
while ($row = mysql_fetch_assoc($queryresult))
{
$array[$i]['title'] = $row['eventTitle'];
$array[$i]['category'] = $row['eventCategory'];
$i++;
}
HTML -
<table border="1">
<tr>
<td>Events</td>
<td>Category</td>
</tr>
<?php
foreach($array as $arr)
{
?>
<tr>
<td><?php echo $arr['event']; ?></td>
<td><?php echo $arr['category']; ?></td>
</tr>
<?php
}
?>
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I output data from atabase in table.
In result table created with next code:
while($i=$res2->fetch_assoc()) {
$a++;
$t2.='
<tr>
<td>'.$a.'</td>
<td>'.date_format(new DateTime($i['date']),'d.m.Y').'</td>
';
if($valid!='id'){
$t2.='
<td>'.$partner.'</td>
';
}
$t2.='
<td>'.$http_referer.'</td>
</tr>';
}
$t1='
<table class="table table-hover table-bordered">
<thead>
<tr>
<th class="column_th_number">№</th>
<th>Date</th>
';
if($valid!='id'){
$t1.='
<th>Partner</th>
';
}
$t1.='
<th></th>
</tr>
<tr>
<td colspane="2">
</td>
</tr>
</thead>
<tbody>
';
$t3='
</tbody>
</table>
';
echo $t1.$t2.$t3;
but in the result I see that last row was not closed (see image):
Tell me please why last row was not closed?
And how can I make this right?
<td>'.date_format(new DateTime($i['date']),'d.m.Y').'</td>
if($valid!='id'){
replace with
<td>'.date_format(new DateTime($i['date']),'d.m.Y').'</td>'; <--- end the string here
if($valid!='id'){
you missed ';
<td>'.$a.'</td>
<td>'.date_format(new DateTime($i['date']),'d.m.Y').'</td>';
if($valid!='id'){
Thanks ALL.
Error was with colspan(value does not match the number of columns).
Thanks all for help
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Just looking for better solution.I'm retrieving all the records from mysql. Then I will send this data by email. Its working fine. Is there any better solution? thanks
$query = "SELECT * FROM order_details WHERE order_id = '".$data['order_id']."'";
while($row = mysql_fetch_array($result,MYSQL_ASSOC)) {
$message_admin .= "<table style='text-align:center;' border='1' cellspacing='0' cellpadding='7'>
<tr>
<td>Source</td>
<td>".$data['source']."</td>
</tr>
<tr>
<td>Email</td>
<td>".$data['email']."</td>
</tr>
<tr>
<td>Message</td>
<td>".$data['message']."</td>
</tr>
<tr>
<td></td>
<td></td>
</tr>
</table>";
}
You would benefit from: http://us.php.net/manual/en/language.types.string.php#language.types.string.syntax.heredoc
Instead of the complicated string manipulation you are doing now.
You are setting $row, but using $data?
I would take a look at the following to improve your code.
mysql-real-escape-string
SELECT * vs SELECT column