I need to get the data of the database to an HTML table.
There were similar questions asked before but none of them were similar to my requirement.
I need to get the first row of the database to the first row of the table.
So I did this.
echo "<table id='main-table' border='1px solid'> <thead> ";
echo "<tr>";
echo "<th>Board No</th>";
echo "<th>Number</th>";
echo "</tr></thead><tbody>";
while($query->fetch()){
echo "<form action='' class='' method='post'><tr>";
line 55 --> echo "<td>" . Board Code:".$row["board_code"] . Number:".$row["status"] . "</td>";
echo "</tr> </form>";
}//End of While
echo "</tbody></table>";
There is a syntax error in line 55 as mentioned above in the code.
Check this code:
echo "<table id='main-table' border='1px solid'> <thead> ";
echo "<tr>";
echo "<th>Board No</th>";
echo "<th>Number</th>";
echo "</tr></thead><tbody>";
while($query->fetch()){
echo "<form action='' class='' method='post'><tr>";
echo "<td> Board Code:".$row["board_code"]. "</td>";
echo "<td>Number:".$row["status"]. "</td>";
echo "</tr> </form>";
}
echo "</tbody></table>";
"<td>" . Board Code:".$row["board_code"] ." Number:".$row["status"] . "</td>";
edit your code like if it does not work please share your whole code will help you for sure
Check with this code
echo "<form action='' class='' method='post'><tr>";
line 55 --> echo "<td>" . Board Code:".$row["board_code"] ." Number:".$row["status"] . "</td>";
echo "</tr> </form>";
Related
i am trying this from 5hrs don't know i am going wrong.My table echos results but not echo total for the same.
<table width="102%" border="0">
<tr>
<th><b>Amount</b></th>
<th ><strong>County</strong></th>
</tr><?php
(this fetches data ...$row = mysqli_fetch_array($res) or die(mysqli_error($db)))
echo "<tr>";
echo "<td>" . $price . "</td>";
echo "<td>" . $rui['county'] . "</td>";
echo "</tr>";
(I get these entreis correct )
}
}
(Below table is not echoed)
echo "<tr>";
echo "<td> .<b>Total </b>. </td>";
echo "<td>" .$x. "</td>";
echo "<td></td>";
echo "</tr>";
?></table>
Remove the } } after the closing tag of the tr.
if you can't close the whole PHP block then you probably did something wrong in your syntax.
I don't know if you have more code or not, if you do - post it so we will get more accurate view of your code.
<table width="102%" border="0">
<tr>
<th><b>Amount</b></th>
<th ><strong>County</strong></th>
</tr>
<?php
(this fetches data ...$row = mysqli_fetch_array($res) or die(mysqli_error($db)))
echo "<tr>";
echo "<td>" . $price . "</td>";
echo "<td>" . $rui['county'] . "</td>";
echo "</tr>";
(I get these entreis correct )
(Below table is not echoed)
echo "<tr>";
echo "<td> .<b>Total </b>. </td>";
echo "<td>" .$x. "</td>";
echo "<td></td>";
echo "</tr>";
}
?>
</table>
You can remove the wrong bracket, OR check that the two bracket are opened when fetching data
<table width="102%" border="0">
<tr>
<th><b>Amount</b></th>
<th ><strong>County</strong></th>
</tr><?php
//(this fetches data ...$row = mysqli_fetch_array($res) or die(mysqli_error($db)))
echo "<tr>";
echo "<td>" . $price . "</td>";
echo "<td>" . $rui['county'] . "</td>";
echo "</tr>";
//(I get these entreis correct )
//(Below table is not echoed)
echo "<tr>";
echo "<td> .<b>Total </b>. </td>";
echo "<td>" .$x. "</td>";
echo "<td></td>";
echo "</tr>";
?>
</table>
I am creating a random generated quiz with 10 questions. In generating the random question is fine, but I want to display also the choices in random, same with the questions.
This is my code that I am currently working with:
<?php
generate();
function generate(){
include('connection.php');
mysql_select_db('exam');
$result=mysql_query("SELECT * FROM questionaires
INNER JOIN choices ON questionaires.q_id=choices.q_id
WHERE RAND()<(SELECT ((10/COUNT(*))*10) FROM questionaires)
ORDER BY RAND() LIMIT 10");
$c=0;
echo "<table border='3' align='center' bordercolor='#CCCCCC'>
<tr>
<th>Number:</th>
<th>Question</th>
</tr>
";
while($row = mysql_fetch_array($result)){
$c++;
echo "<tr>";
echo "<td>" . $c . "</td>";
echo "<td>";
echo $row['question'] . "<br>";
echo "<input type='radio' name='ans'>".$row['choice_a']."</input><br>";
echo "<input type='radio' name='ans'>".$row['choice_b']."</input><br>";
echo "<input type='radio' name='ans'>".$row['choice_c']."</input><br>";
echo "<input type='radio' name='ans'>".$row['choice_d']."</input><br>";
echo "</td>";
echo "</tr>";
//}
//}
}
echo "</table>";
}
?>
Little help will highly appreciated.
You could change script to this:
echo "<td>";
echo $row['question'] . "<br>";
$ans=array($row['choice_a'],$row['choice_b'],$row['choice_c'],$row['choice_d']);
shuffle($ans);
foreach ($ans as $choice) {
echo "<input type='radio' name='ans'>".$choice."</input><br>";
} unset($choice);
echo "</td>";
This is my code (horrible one):
<?php
include 'connect/con.php';
$result = mysqli_query($con,"SELECT id, vidTitle FROM newsvid");
$result1 = mysqli_query($con,"SELECT imgCover, vidSD FROM newsvid");
$result2 = mysqli_query($con,"SELECT published FROM newsvid");
echo "<table width=\"600\" border=\"1\"><tbody>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo '<td width=\"10%\">'.$row['id'].'</td>';
echo "<td width=\"90%\">" . $row['vidTitle'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
echo "<table width=\"600\" border=\"1\"><tbody>";
while($row = mysqli_fetch_array($result1)) {
echo "<tr>";
echo "<td width=\"40%\">" . $row['imgCover'] . "</td>";
echo "<td width=\"60%\">" . $row['vidSD'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
echo "<table width=\"600\" border=\"1\"><tbody>";
while($row = mysqli_fetch_array($result2)) {
echo "<tr>";
echo "<td >" . $row['published'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
mysqli_close($con);
?>
</body>
</html>
The question is how to show data from database in this layout:
--------------------------------
-id----------vidTitle-----------
--------------------------------
-imgCover------vidSD------------
--------------------------------
----------published-------------
So every time I will add more data , another block like I showed before will add up under existing one.
........................................................................................
There's no need to write 3 queries. You could do that with only one select, and then put all the echos inside a while. That way you're writing, it would run all the ids and titles first, then it would put a table after the table, with cover and vidSD.
Try to make a single query:
SELECT id, vidTitle, imgCover, vidSD, published FROM newsvid
That way you will have, on each row returned from database, all the information about the same row.
Now, running a while is the same as you're doing, just adapting some HTML:
echo "<table width='600' border='1'><tbody>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo '<td width=\"10%\">'.$row['id'].'</td>';
echo "<td width=\"90%\">" . $row['vidTitle'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td width=\"40%\">" . $row['imgCover'] . "</td>";
echo "<td width=\"60%\">" . $row['vidSD'] . "</td>";
echo "</tr>";
echo "<tr>";
echo "<td colspan='2'>" . $row['published'] . "</td>";
echo "</tr>";
}
echo "</tbody></table>";
You may want to order it too. Adding ORDER BY id DESC, would do that.
i'm trying to do a table with php which takes the data from my database and build's the table im html but despite my while loop , only the 1st row is getting into the table. If i change the while loop before the
echo "<table border='1' cellpadding='2' cellspacing='2'";
I get all of them but in 3 tables.
What's wrong with the code ?
<?php
require 'connect.php';
$query = $link->query("SELECT * FROM fornecedor");
echo "Fornecedores";
echo "<table border='1' cellpadding='2' cellspacing='2'";
echo "<tr>
<td>Nome</td>
<td>NIF</td>
<td>Cidade</td>
<td>Rua</td>
<td>NrPorta</td>
<td>Website</td>
<td>email</td>
</tr>";
while ($row = mysqli_fetch_array($query)) {
echo "<tr>";
echo "<td>" . $row["Nome"] . "</td>";
echo "<td>" . $row["NIF"] . "</td>";
echo "<td>" . $row["Cidade"] . "</td>";
echo "<td>" . $row["Rua"] . "</td>";
echo "<td>" . $row["NrPorta"] . "</td>";
echo "<td>" . $row["Website"] . "</td>";
echo "<td>" . $row["Email"] . "</td>";
echo "</tr>";
echo "</table>";
};
?>
Move echo "</table>"; out of the while loop-
....
while(){
.....
// don't close the table tag here
}
echo "</table>";
For each row you are adding the closing </table> tag which should not be like that.
echo "</table>"; code must be out of the while loop. It must be written once only.
I am new to PHP and I have problem. I have a table with check boxes. I need to add if statement inside the check box.
echo "<div class='table1'>
<table>
<tr>
<td></td>
<td>Module code</td>
<td>Module Title</td>
<td>Option</td>
</tr>";
echo "<form action='confirmsubmission.php' method='post'>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . "<input type='checkbox' name='check[]'
value='".$row['module_id']."' />" . "</td>";
echo "<td>" . $row['module_id'] . "</td>";
echo "<td>" . $row['module_title'] . "<a href=# content='".$row['description']."'
class='tooltip'><span title='Module Description'><img src='images/i.png'/></span>
</a>". "</td>";
echo "<td>" . $row['module_choice'] . "</td>";
echo "</tr>";
}
echo "</table></div>";
Below is a If statement I need to add after value='".$row['module_id']."'
if($row['module_choice']=='Mandatory'){ echo "checked=\"true\""; }
A bit of a mouthful, but try:
...
echo "<td>" . "<input type='checkbox' name='check[]' value='".$row['module_id']."'".($row['module_choice']=='Mandatory' ? 'checked="true"' : "")." />" . "</td>";
...
Even though I'll advise you to use a templating system (Smarty, etc.), you can do the following code:
echo "<div class='table1'>
<form action='confirmsubmission.php' method='post'>
<table>
<tr>
<td></td>
<td>Module code</td>
<td>Module Title</td>
<td>Option</td>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . "<input type='checkbox' name='check[]'
value='".$row['module_id']."'";
if($row['module_choice']=='Mandatory'){
echo " checked='checked' ";
}
echo "/>" . "</td>";
echo "<td>" . $row['module_id'] . "</td>";
echo "<td>" . $row['module_title'] . "<a href=# content='".$row['description']."'
class='tooltip'><span title='Module Description'><img src='images/i.png'/></span>
</a>". "</td>";
echo "<td>" . $row['module_choice'] . "</td>";
echo "</tr>";
}
echo "</table></form></div>";