ten120963
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Posts posted by ten120963
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Here is all of it:
<?php
define('TITLE', 'Test');
include('templates/header.html');
print '<h2>Test</h2>';
if ($_SERVER['REQUEST_METHOD'] == 'POST') { // handle the form
// need the database connection
include('mysql_connect.php');
// validate the form data
$problem = FALSE;
if (!empty($_POST['name'])) {
// prepare the values for storing:
$name = trim(strip_tags($_POST['name']));
} else {
print '<p style="color: red;">Please submit all fields</p>';
$problem = TRUE;
}
if (!$problem) {
$statusarray = array ('Name' => 0);
// Define the query:
$statusquery = 'SELECT name, status FROM users ORDER BY name';
// run the query
if ($sq = mysql_query($statusquery, $dbc)) {
// retrieve and process every record:
while ($statusrow = mysql_fetch_array($sq)) {
$name = $statusrow['name'];
$status = $statusrow['status'];
$statusarray[$name] = $status;
} // end of fetch
foreach ($statusarray as $aname => $astatus) {
print "<p>$aname: $astatus</p>\n";
}
} else { // Query didn't run.
print '<p style="color: red;">Could not retrieve the data because:<br />' . mysql_error($dbc)
. '.</p><p>The query being run was: ' . $statusquery . '</p>';
} // End of query IF.
$stickytest = '';
if(isset($_POST['submit'])) {
$stickytest = $_POST['name'];
}
print "$stickytest";
print $_POST['name'];
if ($astatus <> 'P') {
print "Invalid choice.";
$valid_choice = 'FALSE';
} else {
$valid_choice = 'TRUE';
}
if ($valid_choice <> 'FALSE') {
// define the query
$query = "INSERT INTO users (id, name) VALUES (0, '$name')";
// execute the query
if (@mysql_query($query, $dbc)) {
print '<p>Name has been added.</p>';
} else {
print '<p style="color: red;">Could not add the name because:<br />' .
mysql_error($dbc) . '.</p><p>The query being run was: ' . $query . '</p>';
}
}
} // no problem
mysql_close($dbc); // close the connection
} // end of form submission if
// display the form
?>
<div><p>Please enter your name:</p>
<form action="add_name.php" method="post">
<table cellpadding="2" cellspacing="2" align="left">
<tr>
<td>Name:</td>
<td><input type="text" name="name" size="20"
value="<?php if (isset($_POST['name'])) { print htmlspecialchars($_POST['name']); } ?>" /></td>
</tr>
<?php
$teams = array ('user1', 'user2');
foreach ($teams as $key => $value) {
print "<tr><td>$value:</td><td><select name=\"$value\">";
include('mysql_connect.php');
$query = "SELECT * FROM $value";
if ($r = mysql_query($query, $dbc)) { // run the query
// if sticky add selected parm
while ($row = mysql_fetch_array($r)) {
if ($stickytest == $row['name']) {
print "HI";
print "<option selected=\"selected\" value=\" {$row['name']} \"> {$row['name']} </option>";
} else {
print "HO";
print "<option value=\" {$row['name']} \"> {$row['name']} </option>";
}
}
} else { // query did not run
print '<p style="color: red;">Could not retrieve the data because:<br />' . mysql_error($dbc)
. '.</p><p>The query being run was: ' . $query . '</p>';
} // end of query if
mysql_close($dbc); // close the connection
print "</select></td></tr>";
}
?>
</table>
<br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />
<input type="submit" name="submit" value="Enter Name" />
</form>
<?php include('templates/footer.html');
?>
</div>
</body>
</html>
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I have seen an example of creating a sticky Drop Down box, but I'm a little confused on how to get it to work.
In the 'if ($_SERVER['REQUEST_METHOD'] == 'POST') { // handle the form' section, I have:
$stickytest = '';
if(isset($_POST['submit'])) {
$stickytest = $_POST['name'];
}
print "$stickytest";
print $_POST['name'];
For this code, the two things that are printed are in fact equal during my testing.
Then, in my '} // end of form submission if' '// display the form' code, I have:
if ($stickytest == $row['name']) {
print "HI";
print "<option selected value=\" {$row['name']} \"> {$row['name']} </option>";
} else {
print "HO";
print "<option value=\" {$row['name']} \"> {$row['name']} </option>";
}
I am expecting to see either HI or HO during my testing, and I am seeing neither.
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Thanks to everyone for the great suggestions. I will try to figure it out from here.
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My code:
$names = array(); // create a variable to hold the query results
$query = "SELECT name FROM users ORDER BY name ASC";
// run the query
if ($r = mysql_query($query, $dbc)) {
// retrieve every record and add to a dropdown box
while ($row = mysql_fetch_array($r, MYSQL_ASSOC)) {
$names[] = $row; // add the row in to the results array
}
// query did not run
} else {
print '<p style="color: red;">Could not retrieve the data because:<br />' . mysql_error($dbc)
. '.</p><p>The query being run was: ' . $query . '</p>';
} // end of query if
mysql_close($dbc); // close the connection
// set up combo box
print '<tr><td>Name:</td><td><select name="name"';
foreach ($names as $num => $tname) {
print "<option value=\"$num\">$tname</option>";
}
print "</select></td>";
I get a dropdown box filled with the word 'Array' several times.
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That could be it, because I do have two of '$row = mysql_fetch_array($r)' in my code. I am trying to create a form that has a table with a pull down menu of these 32 users, then a pull down menu with values in one row, then the same two pull down menus in the row below before i close the table. So, I am using the statement above two times to get the data twice. Would using $row2 and $q to get the data the second time solve the problem, or is there a more efficient way to get the data once to populate 2 different drop downs?
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How would I use print_r in the code above? I tried after the 'while ($row = mysql_fetch_array($r)) {' but did not see anything different on the page.
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It's always the first one that is not shown. If I sort by name, it's the first one alphabetically that is not shown. If I do not sort, then it's the first record in the database that is now shown.
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Sorry, I should have mentioned that in the first post. I did run it thru phpMyAdmin and Sequel Pro, and received all 32 records.
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I have a database table with 32 records in it. I have written PHP code to cycle thru the records, and add the name field from each record into a pull down menu box. So far, I have tried 'select name from users order by name asc' and just 'select name from users', and each time, only 31 of the 32 names are displayed. Has anyone had this happen?
$query = "SELECT name FROM users ORDER BY name ASC";
// run the query
if ($r = mysql_query($query, $dbc)) {
// retrieve every record and add to a dropdown box
while ($row = mysql_fetch_array($r)) {
print "<option value=\" {$row['name']} \"> {$row['name']} </option>";
}
// query did not run
} else {
print '<p style="color: red;">Could not retrieve the data because:<br />' . mysql_error($dbc)
. '.</p><p>The query being run was: ' . $query . '</p>';
} // end of query if
print "</select></td>";
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I have just completed reading PHP For the Web, and just started reading PHP and MYSQL for Dynamic Web Sites. Both are wonderful books. I have a few other PHP/MYSQL books. A few are ok, and a few are terrible. These books are great.
I am only on page 45 of the second book, but I noticed that in the first book, print is used where in the second book, echo is used. Also, in the first book, $_GET is used, where in the second, $_REQUEST is used.
Is there benefits to using echo and $_REQUEST? Please let me know. Thanks.
Sticky Drop Down Box
in PHP for the Web: Visual QuickStart Guide (4th Edition)
Posted
There are no errors being generated. Not only are HI and HO not being outputted, but the choice in the drop down is not sticky, it is reverting back to the first item in the drop down.