Syntax Error in PHP, unexpected '<' [duplicate] - php

Question

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PHP parse/syntax errors; and how to solve them
(20 answers)
Closed 7 years ago.
I have a error problem with my code in php. can anyone help me to correct it!
<?php $query_tbl = "SELECT transaksi.id_transaksi, pelanggan.nama FROM transaksi, pelanggan ".
"WHERE transaksi.id_pelanggan = pelanggan.id_pelanggan ORDER BY id_transaksi";
$list_tbl = mysqli_query($dbc, $query_tbl)
or die('Error select table');
while($row = mysqli_fetch_array($list_tbl))
{
echo '<tr>';
echo '<td>' . $row['id_transaksi'] . '</td>';
echo '<td>' . $row['nama'] . '</td>';
<form method="post" action="pengiriman-input.php">
<input type="hidden" name="id" value="<?php echo $row['id_transaksi']; ?>" />
<input type="submit" value="Kirim" />
</form>
echo '</tr>';
};?>
In Chrome i get this notif:
Parse error: syntax error, unexpected '<' in C:\xampp\htdocs\delivery\pengiriman.php on line 33.
I don't know what is wrong because i am new with php.
And btw line 33 is
<form method="post" action="pengiriman-input.php">

Answer 1

there is some syntex error while putting HTML between PHP near your form tag
<?php $query_tbl = "SELECT `transaksi.id_transaksi`, `pelanggan.nama` FROM `transaksi`, `pelanggan` WHERE
`transaksi.id_pelanggan` = `pelanggan.id_pelanggan` ORDER BY `id_transaksi`";
$list_tbl = mysqli_query($dbc, $query_tbl)
or die('Error select table');
while($row = mysqli_fetch_array($list_tbl))
{
echo '<tr>';
echo '<td>' . $row['id_transaksi'] . '</td>';
echo '<td>' . $row['nama'] . '</td>';?>
<form method="post" action="pengiriman-input.php">
<input type="hidden" name="id" value="<?php echo $row['id_transaksi']; ?>" />
<input type="submit" value="Kirim" />
</form>
<?php echo '</tr>';
};?>

Answer 2

Try this more simple and neat code
<?php
$query_tbl = "SELECT transaksi.id_transaksi, pelanggan.nama FROM transaksi, pelanggan ". "WHERE transaksi.id_pelanggan = pelanggan.id_pelanggan ORDER BY id_transaksi";
$list_tbl = mysqli_query($dbc, $query_tbl) or die('Error select table');
?>
<?php while($row = mysqli_fetch_array($list_tbl)): ?>
<tr>
<td><?php echo $row['id_transaksi']; ?></td>
<td><?php echo $row['nama']; ?></td>
<td>
<form method="post" action="pengiriman-input.php">
<input type="hidden" name="id" value="<?php echo $row['id_transaksi']; ?>" />
<input type="submit" value="Kirim" />
</form>
</td>
</tr>
<?php endwhile; ?>

Answer 3

You can't put HTML directly into PHP
<?php $query_tbl = "SELECT transaksi.id_transaksi, pelanggan.nama FROM transaksi, pelanggan ".
"WHERE transaksi.id_pelanggan = pelanggan.id_pelanggan ORDER BY id_transaksi";
$list_tbl = mysqli_query($dbc, $query_tbl)
or die('Error select table');
while($row = mysqli_fetch_array($list_tbl))
{
echo '<tr>';
echo '<td>' . $row['id_transaksi'] . '</td>';
echo '<td>' . $row['nama'] . '</td>';
?> <!-- here ends the PHP -->
<form method="post" action="pengiriman-input.php">
<input type="hidden" name="id" value="<?php echo $row['id_transaksi']; ?>" />
<input type="submit" value="Kirim" />
</form>
<?php // here begin PHP again
echo '</tr>';
};?>

Answer 4

You need to end the php then start it again. Try this:
<?php $query_tbl = "SELECT transaksi.id_transaksi, pelanggan.nama FROM transaksi, pelanggan ".
"WHERE transaksi.id_pelanggan = pelanggan.id_pelanggan ORDER BY id_transaksi";
$list_tbl = mysqli_query($dbc, $query_tbl)
or die('Error select table');
while($row = mysqli_fetch_array($list_tbl))
{
echo '<tr>';
echo '<td>' . $row['id_transaksi'] . '</td>';
echo '<td>' . $row['nama'] . '</td>';
?>
<form method="post" action="pengiriman-input.php">
<input type="hidden" name="id" value="<?php echo $row['id_transaksi']; ?>" />
<input type="submit" value="Kirim" />
</form>
<?php
echo '</tr>';
};?>

Answer 5

In the above snippet you are opening a php tag in line 1 and have thought of closing it at last line. Between these two tag you cannot write the html script for creating form, as you are doing, in a php file.
It will be taken as a php script instead of html script and php compiler don't understands html tags.
Create the form tag as a php string variable and echo it as you are doing in the lines above.