print records in table format

Question

himanshumishra 0 Newbie Poster

13 Years Ago

Dear All,

I wish to print my records of following code in table format

echo "LOAN NO =  $title " ;
echo "NAME =  $title1 " ;
echo "ADDRESS =  $title2 " ;
echo "AREA OFFICE CODE =  $title3 " ;
echo "LINK AO =  $title4 " ;
echo "APPLICATION NO. =  $title5 " ;
echo "INTREST RATE =  $title6 " ;
echo "CURRENT OS =  $title7 " ;
echo "FUP =  $title8 " ;

I wish the output should come as 1 9*2 table with variable name and values.

Please modify the code for the same

php

2 Contributors
8 Replies
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6 Days Discussion Span
Latest Post 13 Years Ago Latest Post by himanshumishra

Shanti C 106 Posting Virtuoso

13 Years Ago

<table border=1>
<tr>
<td>LOAN NO</td>
<td><?php=$title?></td>
</tr>
<tr>
<td>NAME</td>
<td><?php=$title1?></td>
</tr>
<tr>
<td>ADDRESS</td>
<td><?php=$title2?></td>
</tr>
.
.
.
.etc
</table>

Shanti C 106 Posting Virtuoso

13 Years Ago

then post your code...

Don't put the above code in between php tags.

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himanshumishra 0 Newbie Poster · Answer 1 · 2011-01-25T13:51:28+00:00

it gives Parse error: parse error in C:\wamp\www\search.php on line "<table border=1>"

himanshumishra 0 Newbie Poster · Answer 2 · 2011-01-31T13:22:40+00:00

Dear Shanti,

i have a very basic knowledge of PHP . I am still getting the parse error, can u please modify in my existing code.

thanking in advance

======
<html>

<head>
<meta http-equiv="Content-Type" content="text/html; charset=windows-1252">
<title>Search Result</title>
</head>

<?php

// Get the search variable from URL

$var = @$_GET ;
$trimmed = trim($var); //trim whitespace from the stored variable

// rows to return
$limit=10;

// check for an empty string and display a message.
if ($trimmed == "")
{
echo "Please enter a search...";
exit;
}

// check for a search parameter
if (!isset($var))
{
echo "We dont seem to have a search parameter!";
exit;
}

//connect to your database ** EDIT REQUIRED HERE **
mysql_connect("localhost","root","admin123"); //(host, username, password)

//specify database ** EDIT REQUIRED HERE **
mysql_select_db("mis") or die("Unable to select database"); //select which database we're using

$field = 'lnno';
// Build SQL Query
$query = "select * from loandata where $field = $trimmed
order by $field"; // EDIT HERE and specify your table and field names for the SQL query

$numresults=mysql_query($query);
$numrows=mysql_num_rows($numresults);

// If we have no results,

if ($numrows == 0)
{

echo "Sorry, your search: "" . $trimmed . "" returned zero results";
exit;

}

// next determine if s has been passed to script, if not use 0
if (empty($s)) {
$s=0;
}

// get results
$query .= " limit $s,$limit";
$result = mysql_query($query) or die("Couldn't execute query");

// display what the person searched for
echo "You searched for LOAN NO: "" . $var . """;

// begin to show results set
echo "<h3>Results</h3>";
$count = 1 + $s ;

// now you can display the results returned
while ($row= mysql_fetch_array($result)) {
$title = $row["$field"];
$title1 = $row["mname"];
$title2 = $row["address"];
$title3 = $row["area_off"];
$title4 = $row["link_ao"];
$title5 = $row["appl_No"];
$title6 = $row["int_rate"];
$title7 = $row["curr_os"];
$title8 = $row["fup"];
$title9 = $row["area_name"];

echo "LOAN NO =  $title " ;
echo "NAME =  $title1 " ;
echo "ADDRESS =  $title2 " ;
echo "AREA OFFICE CODE =  $title3 " ;
echo "LINK AO =  $title4 " ;
echo "APPLICATION NO. =  $title5 " ;
echo "INTREST RATE =  $title6 " ;
echo "CURRENT OS =  $title7 " ;
echo "FUP =  $title8 " ;
echo "AREA OFFICE NAME =  $title9 " ;

$count++ ;
}

$currPage = (($s/$limit) + 1);

//break before paging
echo " ";

?>

</body>

</html>

Shanti C 106 Posting Virtuoso · Answer 3 · 2011-01-31T13:31:29+00:00

i checked your code it is perfectly running..
Please post your error and code line numbers.

himanshumishra 0 Newbie Poster · Answer 4 · 2011-01-31T13:52:11+00:00

I need to print the following output in table format.Kindly modify this existin code for the same.

--

echo "LOAN NO =  $title " ;
echo "NAME =  $title1 " ;
echo "ADDRESS =  $title2 " ;
echo "AREA OFFICE CODE =  $title3 " ;
echo "LINK AO =  $title4 " ;
echo "APPLICATION NO. =  $title5 " ;
echo "INTREST RATE =  $title6 " ;
echo "CURRENT OS =  $title7 " ;
echo "FUP =  $title8 " ;
echo "AREA OFFICE NAME =  $title9 " ;

----

Shanti C 106 Posting Virtuoso · Answer 5 · 2011-01-31T14:04:53+00:00

check this code:

echo "<table border=5 cellpadding=2 cellspacing=1>";
// now you can display the results returned
while ($row= mysql_fetch_array($result)) {
$title = $row["$field"];
$title1 = $row["mname"];
$title2 = $row["address"];
$title3 = $row["area_off"];
$title4 = $row["link_ao"];
$title5 = $row["appl_No"];
$title6 = $row["int_rate"];
$title7 = $row["curr_os"];
$title8 = $row["fup"];
$title9 = $row["area_name"];

echo "<tr><td>";
echo "LOAN NO </td><td> &nbsp;$title <br>" ;
echo "</tr></td>";
echo "<tr><td>";
echo "NAME </td><td> &nbsp;$title1 <br>" ;
echo "</tr></td>";
echo "<tr><td>";
echo "ADDRESS </td><td> &nbsp;$title2 <br>" ;
echo "</tr></td>";
echo "<tr><td>";
echo "AREA OFFICE CODE </td><td> &nbsp;$title3 <br>" ;
echo "</tr></td>";
echo "<tr><td>";
echo "LINK AO </td><td> &nbsp;$title4 <br>" ;
echo "</tr></td>";
echo "<tr><td>";
echo "APPLICATION NO. </td><td> &nbsp;$title5 <br>" ;
echo "</tr></td>";
echo "<tr><td>";
echo "INTREST RATE </td><td> &nbsp;$title6 <br>" ;
echo "</tr></td>";
echo "<tr><td>";
echo "CURRENT OS </td><td> &nbsp;$title7 <br>" ;
echo "</tr></td>";
echo "<tr><td>";
echo "FUP </td><td> &nbsp;$title8 <br>" ;
echo "</tr></td>";
echo "<tr><td>";
echo "AREA OFFICE NAME </td><td> &nbsp;$title9 <br>" ;

$count++ ;
}
echo "</table>";

himanshumishra 0 Newbie Poster · Answer 6 · 2011-01-31T14:56:25+00:00

thank you very much .... it worked perfectly !!!!!!!!!!!!