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I am trying to output a list of links based upon database query result. The query works fine but I need to get the information for the links to populated by the same results. In this example I need to create the following string to pass along to my loadXMLDoc() Function..
$ridesid
$catrating
This is the actual line of code that is giving me an issue:
echo "<a href='#' onmousedown='url1 ='attraction_page.php?rideid=$ridesid&catrating=$catrating onclick ='loadXMLDoc()'>$rides</a>";
I am pretty sure it is just an issue with the quotes but I have been searching and searching for a solution or an example of how to properly code this line but can't find anything. Can anyone help?
Thanks
Here is the complete PHP block:
<?php
$sql3 = "SELECT `Record_ID`, `Name` FROM `rides` WHERE `Rating` = $catrating ORDER BY `Name`";
$result3=mysql_query($sql3)or die(mysql_error());
//var_dump ($result3);
$num = mysql_num_rows($result3);
//WHILE ($row3 = mysql_fetch_array($result3)){
echo "<table>";
// Ride Category Heading
if ($catrating == 1)
echo "<span class='headertext'>Kiddie Rides</span><br \>";
if ($catrating == 2)
echo "<span class='headertext'>Family Rides</span><br \>";
if ($catrating == 3)
echo "<span class='headertext'>Thrill Rides</span><br \>";
// end heading
for ($i = 0; $i < $num; $i++){
$row3 = mysql_fetch_array($result3);
//var_dump($row3);
$ridesid = $row3[0];
$rides = $row3[1];
//echo "<a href='attraction_page.php?rideID=". urlencode($ridesid) ."&catrating=". urlencode($catrating) ."'>$rides</a>";
echo "<tr>";
//echo "<a href='#' onmousedown='url1 ='attraction_page.php?rideID=$ridesid&catrating=$catrating' onclick='loadXMLDoc()'>$rides</a>";
//echo "<a href='#' onmousedown='url1 ='attraction_page.php?rideID=30&catrating=1 onclick='loadXMLDoc()'>$rides</a>";
//echo "<a href='attraction_page.php?rideID=". urlencode($ridesid) ."&catrating=". urlencode($catrating) ."'>$rides</a>";
echo "<a href='#' onmousedown='url1 ='attraction_page.php?rideid=$ridesid&catrating=$catrating onclick ='loadXMLDoc()'>$rides</a>";
echo "<br />";
echo "</tr>";
}
echo '</table>';
// }
?>
You must escape your quotes.
echo "<a href='javascript:void(0);' onmousedown='url1 =\'attraction_page.php?rideid=$ridesid&catrating=$catrating\';' onclick ='loadXMLDoc()'>$rides</a>";
I would also replace the # in your href attribute with javascript:void(0)
I would recommend to leave quotes when you want to use a var, or use the {$var} syntax
echo "<a href='#' onmousedown='url1 =\'attraction_page.php?rideid=".$ridesid."&catrating=".$catrating."\'' onclick ='loadXMLDoc()'>".$rides."</a>";
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Why does this not work? I want to be redirected to id?1 and remove.php removes it from table. How is this done?
// Get all the data from the "example" table
$result = mysql_query("SELECT * FROM pages")
or die(mysql_error());
echo "<table border='1'>";
while ($row = mysql_fetch_array($result)) {
echo "<li class='list-group-item'>";
echo $row['header'];
echo "<br/>";
echo $row['description'];
echo "<br/>";
echo "<button type='button' class='btn btn-default btn-xs' style='margin-top:8px;margin-bottom:0px;'>Ändra innehåll</button>";
echo " ";
echo "<form method='POST'><a href='delete.php?". $row['page_id']."' name='action' class='btn btn-danger btn-xs' style='margin-top:8px;margin-bottom:0px;'>Ta bort sida</a></form>";
echo "<td><a href='obrisi.php?id=$id'>Delete</a></td>";
echo "</li>";
}
echo "</table>";
?>
delete.php
<?php
if (isset($_POST['1']))
{
foreach ($_POST['id'] as $page_id)
{
if (!mysql_query("DELETE FROM pages WHERE id = '$page_id'"))
{
echo mysql_error();
}
}
}
?>
Please help me. Thanks in advance.
You can not get the id in POST becuase you are using it as a Query String. You can get it as:
$page_id = intval( $_GET['id'] );
Now you can check it as:
if( $page_id > 0 ){
// your stuff.. use this id in your query
}
What I have changed:
Use $_GET (SUPER GLOBAL) instead of $_POST becuase of query string.
Side Note:
There is no need to use foreach loop becuase it's not an array.
I suggest you to use mysqli_* or PDO instead of mysql_* becuase its deprecated and not available in PHP 7.
Form - PHP Manual
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What´s wrong with the following code?
$cores = array ("#FF0000","#FFBF00","#FFFF00","#04B404","#58FAF4","#0101DF");
foreach ($cores as $cor)
{
echo "<tr>";
echo "<td bgcolor = $cor></td>";
echo "</tr>";
}
But this code works:
$cores = array ("#FF0000","#FFBF00","#FFFF00","#04B404","#58FAF4","#0101DF");
for ($i=0; $i<7; $i++)
{
echo "<tr>";
echo "<td bgcolor=$cores[$i]></td>";
echo "</tr>";
}
Besides, it is not giving the colors in columns (which is the goal), but in rows.
You are missing quotes around the variable:
foreach ($cores as $cor)
{
echo "<tr>";
echo "<td bgcolor = '$cor'></td>";
echo "</tr>";
}
Do this:
$colors = array("#FF0000","#FFBF00","#FFFF00","#04B404","#58FAF4","#0101DF");
$color = $colors[array_rand($colors)];
and pass $color variable to bgcolor in td
echo "<tr>";
echo "<td bgcolor = '$color'></td>";
echo "</tr>";
and it will pick up colors randomly.
If your goal is the column, use:
<?php
$cores = array ("#FF0000","#FFBF00","#FFFF00","#04B404","#58FAF4","#0101DF");
echo "<table>";
echo "<tr>";
foreach ($cores as $cor)
{
echo "<td bgcolor= \"$cor\">1</td>";
}
echo "</tr>";
echo "</table>";
?>
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i want to apply css for a while looping but it returns only the first data if i use css , and without css it returns all the data normally. here is my code it's simple.
$users=mysql_query("select * from users");
while($user=mysql_fetch_array($users))
{
echo "<table>";
echo "<tr>";
echo "<td class='user'>";
echo $user['pseudo'];
echo "</tr>";
echo "</td>";
echo "</table>";
}
Your problem is position:fixed. This tells the browser to lay all tables, one over the other, at window position right:500; left:500; top:100;. So, everything is there, but you tell the browser to put each new data set into a new table, and stack it in front of or behind the previous dataset (=table).
My best guess as to what you wanted to achieve is
$users=mysql_query("select * from users");
echo "<table>";
while($user=mysql_fetch_array($users))
{
echo "<tr>";
echo "<td class='user'>";
echo $user['pseudo'];
echo "</td>";
echo "</tr>";
}
echo "</table>";
$users=mysql_query("select * from users");
echo "<table>";
while($user=mysql_fetch_array($users))
{
echo "<tr>";
echo "<td class='user'>";
echo $user['pseudo'];
echo "</td>";
echo "</tr>";
}
echo "</table>";
Your HTML was malformed before. The tr was closed, before the td was (<tr><td></tr></td>) which is invalid. Also it makes more sense to wrap the loop in the table, instead of declaring it in every iteration.
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I have a simple code which gives me error. Could someone help me to explain what would be the best solution. I am not just interested the solution rather than the explanation as I have been struggling with this type of listing. I have googled it but I could not find a dummy explanation.
Many thanks
<?php
// Run a Query
$result = mysql_query("SELECT * FROM weblinks ORDER BY yeargroup ASC");
$a=-1;
while ($row = mysql_fetch_array($result)) {
if ($row['yeargroup'] == $a) {
echo "<h2 class=\"class\">Year ".$row['yeargroup']."</h2>";
} else {
echo "<ul><li>";
echo "<img src=\"" . $row['img'] . "\"/>";
echo "".$row['webname']." - ".$row['weblink'];
echo "<div class=\"text\">".$row['content']."</div>";
echo "</li></ul>";
}
$a=$row['yeargroup'];
}
?>
fixed the typo error, thanks
This is how it should be
Year 1
Link1
Link2
Link3
Year 2
Link1
Link2
Link3
etc.
$a = null;
while ($row = mysql_fetch_array($result)) {
//check if $a is equal to the current yeargroup
//if not, output the header and open the list
if ($row['yeargroup'] <> $a) {
echo "<h2 class=\"class\">Year ".$row['yeargroup']."</h2>";
echo "<ul>";
}
echo "<li>";
echo "<img src=\"" . $row['img'] . "\"/>";
echo "".$row['webname']." - ".$row['weblink'];
echo "<div class=\"text\">".$row['content']."</div>";
echo "</li>";
//check if $a is equal to the current yeargroup
//if not, close the list and set $a to the yeargroup for the loop
if ($row['yeargroup'] <> $a) {
echo "</ul>";
$a = $row['yeargroup'];
}
}
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New to php. Trying to create a list of images with text in a loop. The script works fine with just , but I want to add a class name to it so that I can control it with css, but I am getting an error. Seems not to flow cleanly with HTML. Any suggestions?
$rs = $conn->Execute("SELECT ProductName, ProductID FROM Products ");
//opens a recordset from the connection object
while (!$rs->EOF) {
$fv1=$rs->Fields("ProductID");
$fv2=$rs->Fields("ProductName");
print "<div class="product-img"> ";
print "<a href='moredetails.php?productid=$fv1'>";
print "<img src = \"thumbs/artifact-$fv1.jpg\" alt = \"Artifact-$fv1\"</a><br>";
print "<a href='moredetails.php?productid=$fv1'>";
print "$fv2</a><br>";
print "</div>";
$rs->MoveNext();
}
$rs->Close();
The double quotation will close the php function, try to use a single quotation for your class attr.
print "<div class='product-img'> ";
or escape it
print "<div class=\"product-img\"> ";
Change
print "<div class="product-img"> ";
to
print "<div class=\"product-img\"> ";