[SOLVED] Send results from one php script to another script

mjgdunne · April 23, 2008

Hi, i have a basic search working and i want to add an edit button so that i can send the results from the search to another php script so that i can edit the details, is this possible, im new to php thanks.

April 23, 2008

<?php session_start();
$_SESSION['varname'] = "variable contents";?>

You have to start the session on every page you want to access the info.

I would recommend doing a $varname = $_SESSION['varname'] on the pages you need to access the info. This makes it easier to use the data.

mjgdunne · April 23, 2008

Hi, i have the following code:

if (isset($_POST['myq'])) {
    $myq = mysql_real_escape_string($_POST['myq']);
    $result = mysql_query("SELECT * FROM members WHERE Firstname LIKE '%$myq' OR Surname LIKE '%$myq'");

    while ($row = mysql_fetch_assoc($result)) {

echo "<p>You searched for: "" . $myq . ""</p>";

So would i put $result in like: $_SESSION['$results']= "results";

April 23, 2008

Actually, you would do $_SESSSION['results'] = $results; That will store the query in the session variable.

I may have confused on the last part of my message. When you need to access the info on another page, you do

<?php
// It is a good habit to start the session at the top of the page
session_start();
$results = $_SESSION['results'];

// Process the data
?>

mjgdunne · April 23, 2008

Thanks, does this mean i can access the results in the session on any php script, and would i just use session_start(); at the start of each script.

April 23, 2008

Exactly. The session can be used as long as the browser window is open, no matter what page you are on.

mjgdunne · April 23, 2008

If i use exit(); at the end if the script does that not exit the session, sorry these questions are prop very basic, thanks.

April 23, 2008

That will not end the session. There are only 2 ways to end the session: session_destroy() and some other way I cannot remember.

EDIT: The "other way" is session_unset(), which only clears the variables; it does not end the session.

mjgdunne · April 23, 2008

Thanks, but im getting an error:

Warning: session_start() [function.session-start]: Cannot send session cache limiter - headers already sent (output started at C:\xampp\htdocs\search_emp.php:15) in C:\xampp\htdocs\search_emp.php on line 30

Any idea what it could be?

April 23, 2008

Move the session_start() to the very top of your script.

mjgdunne · April 23, 2008

Here is my code:

<html>
<head>
<title>Car Rentals & Returns</title>
<meta http-equiv="Content-Type" content="text/html" />
<link href="style2.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="wrapper3">
<img src="images/cars.jpg" width="996" height="100"></a>

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<td align="center">
<H1>Search Results:</H1>
<form action="edit_emp.php" method="post">
</td>
</table>

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=30%>
<tr>
<TD ALIGN=CENTER VALIGN=TOP WIDTH=50>
</TD>
<TD ALIGN=LEFT VALIGN=TOP WIDTH=83%>
<TR><TD>



<?php
session_start();

$_SESSION['$result'] = "variable contents";

ini_set( 'display_errors', '1' );
error_reporting ( 2047 );


$host="localhost"; // Host name
$username="root"; // Mysql username
$password="root"; // Mysql password
$db_name="test"; // Database name
$tbl_name="members"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "root", "root")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

$myq=$_POST['myq'];



if (isset($_POST['myq'])) {
    $myq = mysql_real_escape_string($_POST['myq']);
    $result = mysql_query("SELECT * FROM members WHERE Firstname LIKE '%$myq' OR Surname LIKE '%$myq'");

    while ($row = mysql_fetch_assoc($result)) {

echo "<p>You searched for: "" . $myq . ""</p>";


echo "<table border=\"1\" align=\"center\">";

echo "<tr><th>First Name</th>";
	echo "<td>";
	echo $row['Firstname'];
echo "</td>";

echo "<tr><th>Surname</th>";
	echo "<td>";
	echo $row['Surname'];
echo "</td>";

echo "<tr><th>Username</th>";
	echo "<td>";
	echo $row['username'];
echo "</td>";

echo "<tr><th>Password</th>";
	echo "<td>";
	echo $row['password'];
echo "</td>";

echo "<tr><th>Email</th>";
	echo "<td>";
	echo $row['email'];
echo "</td>";

echo "<tr><th>Level</th>";
	echo "<td>";
	echo $row['level'];
echo "</td>";



    }
}

if (!$result) {
    $message  = 'Invalid query: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query;
    die($message);
}

if (mysql_num_rows($result) == 0) {
    echo "No records found";
    exit;
}



exit();
?>


<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<input type="submit" value="Update Details"/>
</form>
<form method="post" action="login_success3.php">
<td align="center"><input type="submit" value="Home"/></form>
</td>
</tr>
</table>
</table>
</td>
</tr>
</table>
</BODY>
</HTML>

I am still getting the same error.

April 23, 2008

<?php
//Added this session_start()
session_start();
?>
<html>
<head>
<title>Car Rentals & Returns</title>
<meta http-equiv="Content-Type" content="text/html" />
<link href="style2.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="wrapper3">
<img src="images/cars.jpg" width="996" height="100"></a>

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<td align="center">
<H1>Search Results:</H1>
<form action="edit_emp.php" method="post">
</td>
</table>

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=30%>
<tr>
<TD ALIGN=CENTER VALIGN=TOP WIDTH=50>
</TD>
<TD ALIGN=LEFT VALIGN=TOP WIDTH=83%>
<TR><TD>



<?php
// Got rid of this session

// Moved this session declaration

ini_set( 'display_errors', '1' );
error_reporting ( 2047 );


$host="localhost"; // Host name
$username="root"; // Mysql username
$password="root"; // Mysql password
$db_name="test"; // Database name
$tbl_name="members"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "root", "root")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

$myq=$_POST['myq'];



if (isset($_POST['myq'])) {
    $myq = mysql_real_escape_string($_POST['myq']);
    $result = mysql_query("SELECT * FROM members WHERE Firstname LIKE '%$myq' OR Surname LIKE '%$myq'");
// There was nothing to declare before
$_SESSION['result'] = $result;


    while ($row = mysql_fetch_assoc($result)) {

echo "<p>You searched for: "" . $myq . ""</p>";


echo "<table border=\"1\" align=\"center\">";

echo "<tr><th>First Name</th>";
	echo "<td>";
	echo $row['Firstname'];
echo "</td>";

echo "<tr><th>Surname</th>";
	echo "<td>";
	echo $row['Surname'];
echo "</td>";

echo "<tr><th>Username</th>";
	echo "<td>";
	echo $row['username'];
echo "</td>";

echo "<tr><th>Password</th>";
	echo "<td>";
	echo $row['password'];
echo "</td>";

echo "<tr><th>Email</th>";
	echo "<td>";
	echo $row['email'];
echo "</td>";

echo "<tr><th>Level</th>";
	echo "<td>";
	echo $row['level'];
echo "</td>";



    }
}

if (!$result) {
    $message  = 'Invalid query: ' . mysql_error() . "\n";
    $message .= 'Whole query: ' . $query;
    die($message);
}

if (mysql_num_rows($result) == 0) {
    echo "No records found";
    exit;
}



exit();
?>


<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<input type="submit" value="Update Details"/>
</form>
<form method="post" action="login_success3.php">
<td align="center"><input type="submit" value="Home"/></form>
</td>
</tr>
</table>
</table>
</td>
</tr>
</table>
</BODY>
</HTML>

Hope that works.

mjgdunne · April 23, 2008

Thanks, that got rid of the error, if i use the following displaying the results in input fields will it work:

<?php
session_start();
?>

<html>
<head>
<title>Car Rentals & Returns</title>
<meta http-equiv="Content-Type" content="text/html" />
<link href="style2.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="wrapper3">
<img src="images/cars.jpg" width="996" height="100"></a>

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<td align="center">
<H1>Search Results:</H1>
<form action="edit_emp.php" method="post">
</td>
</table>

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=30%>
<tr>
<TD ALIGN=CENTER VALIGN=TOP WIDTH=50>
</TD>
<TD ALIGN=LEFT VALIGN=TOP WIDTH=83%>
<TR><TD>


<?php

$_SESSION['result'] = "$result";

ini_set( 'display_errors', '1' );
error_reporting ( 2047 );


$host="localhost"; // Host name
$username="root"; // Mysql username
$password="root"; // Mysql password
$db_name="test"; // Database name
$tbl_name="members"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "root", "root")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");

$myq=$_POST['myq'];


while ($row = mysql_fetch_assoc($result)) {

echo "<p>You searched for: "" . $myq . ""</p>";


echo "<table border=\"1\" align=\"center\">";

echo "<tr><th>First Name</th>";
	echo "<td>";
	echo '<input type="text" name="firstname" value="'. $row['Firstname'] .'" size="20" />';
echo "</td>";

echo "<tr><th>Surname</th>";
	echo "<td>";
	echo '<input type="text" name="surname" value="'. $row['Surname'] .'" size="20" />';
echo "</td>";

echo "<tr><th>Username</th>";
	echo "<td>";
	echo '<input type="text" name="username" value="'. $row['username'] .'" size="20" />';
echo "</td>";

echo "<tr><th>Password</th>";
	echo "<td>";
	echo '<input type="text" name="password" value="'. $row['password'] .'" size="20" />';;
echo "</td>";

echo "<tr><th>Email</th>";
	echo "<td>";
	echo '<input type="text" name="email" value="'. $row['email'] .'" size="20" />';
echo "</td>";

echo "<tr><th>Level</th>";
	echo "<td>";
	echo '<input type="text" name="level" value="'. $row['level'] .'" size="20" />';
echo "</td>";



    }
}





exit();
?>


<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<input type="submit" value="Update Details"/>
</form>
<form method="post" action="login_success3.php">
<td align="center"><input type="submit" value="Home"/></form>
</td>
</tr>
</table>
</table>
</td>
</tr>
</table>
</BODY>
</HTML>

April 23, 2008

I do not see why id would not. Did you try it and it did not work?

mjgdunne · April 23, 2008

Hi, my button to go to the update page has dissapeared, and is not showing up along with the home button?

April 23, 2008

If you use exit(), it will stop right there and not read anything else. Move the html code before the exit().

mjgdunne · April 23, 2008

Thanks that fixed that, it is having a problem with the line:

while ($row = mysql_fetch_assoc($result)) {

Is there any other way i could just take $result so i can extract the name etc.

Thanks.

April 23, 2008

If you have one row of data:

$row = mysql_fetch_array($result);

Then, reference the values by columname.

echo "Name:".$row['name'];

mjgdunne · April 23, 2008

Thanks, i put that in, but two errors, any idea?

Notice: Undefined variable: result in C:\xampp\htdocs\edit_emp.php on line 52

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\edit_emp.php on line 52

April 23, 2008

Did you replace the while statement with the the $row = mysql_fetch_array($result)? Or did you just add it? (Do not forget to delete the } at the end)

mjgdunne · April 23, 2008

I just replaced it, should i leave it in? Here is the code:

<?php
session_start();
?>

<html>
<head>
<title>Car Rentals & Returns</title>
<meta http-equiv="Content-Type" content="text/html" />
<link href="style2.css" rel="stylesheet" type="text/css" />
</head>
<body>
<div id="wrapper3">
<img src="images/cars.jpg" width="996" height="100"></a>

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<td align="center">
<H1>Search Results:</H1>
<form action="edit_emp.php" method="post">
</td>
</table>

<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=30%>
<tr>
<TD ALIGN=CENTER VALIGN=TOP WIDTH=50>
</TD>
<TD ALIGN=LEFT VALIGN=TOP WIDTH=83%>
<TR><TD>


<?php

$_SESSION['result'] = "$result";

ini_set( 'display_errors', '1' );
error_reporting ( 2047 );


$host="localhost"; // Host name
$username="root"; // Mysql username
$password="root"; // Mysql password
$db_name="test"; // Database name
$tbl_name="members"; // Table name

// Connect to server and select databse.
mysql_connect("$host", "root", "root")or die("cannot connect");
mysql_select_db("$db_name")or die("cannot select DB");




$row = mysql_fetch_array($result);



echo "<table border=\"1\" align=\"center\">";

echo "<tr><th>First Name</th>";
	echo "<td>";
	echo '<input type="text" name="firstname" value="'. $row['Firstname'] .'" size="20" />';
echo "</td>";

echo "<tr><th>Surname</th>";
	echo "<td>";
	echo '<input type="text" name="surname" value="'. $row['Surname'] .'" size="20" />';
echo "</td>";

echo "<tr><th>Username</th>";
	echo "<td>";
	echo '<input type="text" name="username" value="'. $row['username'] .'" size="20" />';
echo "</td>";

echo "<tr><th>Password</th>";
	echo "<td>";
	echo '<input type="text" name="password" value="'. $row['password'] .'" size="20" />';;
echo "</td>";

echo "<tr><th>Email</th>";
	echo "<td>";
	echo '<input type="text" name="email" value="'. $row['email'] .'" size="20" />';
echo "</td>";

echo "<tr><th>Level</th>";
	echo "<td>";
	echo '<input type="text" name="level" value="'. $row['level'] .'" size="20" />';
echo "</td>";



exit();
?>


<TABLE BGCOLOR="#F0F8FF" BORDER=0 CELLPADDING=10 WIDTH=100%>
<tr>
<input type="submit" value="Update Details"/>
</form>
<form method="post" action="login_success3.php">
<td align="center"><input type="submit" value="Home"/></form>
</td>
</tr>
</table>
</table>
</td>
</tr>
</table>
</BODY>
</HTML>

April 23, 2008

Uh oh...I just realized you do not have a query. You have to query the database before you can use information.

mjgdunne · April 23, 2008

I used

if (isset($_POST['myq'])) {
    $myq = mysql_real_escape_string($_POST['myq']);
    $result = mysql_query("SELECT * FROM members WHERE Firstname LIKE '%$myq' OR Surname LIKE '%$myq'");

This is the edit results page, i need to get the results from results, do i still need a query?

April 23, 2008

Whoa, I was all confused there for a minute.

You have to use the previously established $_SESSION['variable'] to access the information.

Use $row = mysql_fetch_array($_SESSION['result']). It should work, but I am not sure.

mjgdunne · April 23, 2008

Thanks for all your help so far, only one error left now,

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in C:\xampp\htdocs\edit_emp.php on line 51

Sign In

[SOLVED] Send results from one php script to another script

Recommended Posts

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Link to comment

Share on other sites

Join the conversation

Important Information