I'm trying to display an image file from a directory using a PHP echo command and an IMG tag.
Here is the code:
//These variables represent the file name extensions from a form element from a previous page
$bannerimg=$_POST["banimg"];
$adimage=$_POST["adimage"];
echo "<img src='imgdir/'".$bannerimg."/>";
When I echo out the file variables ($bannerimg and $adimage) I get the proper file name and extension.
In theory, will this work? If so, what is the proper syntax to handle that echo statement?
Thanks for all the help.
Dustin
Yes it would work, but you should have tested that already.
Alternative syntaxes to the echo statement that I find a little bit more readable would be:
echo "<img src='imgdir/{$bannerimg}/>";
echo "<img src='imgdir/$bannerimg/>";
You can read all about variable parsing in the manual, the first syntax is the complex one and the second the simple. I prefer the complex one as the end of the variable is clearly defined and you can use it for complex expressions, not just simple variables.
You're doing it right but you can use the following just to keep it clean.
echo "<img src='imgdir/$bannerimg' />";
It should work.
To Sandeep's comment, I would have gone the other way.
echo '<img src="imgdir/'.$bannerimg.'/>';
Using " means the parser needs to check to see if there is anything to evaluate.
Related
Please help me with this problem.
<?php echo $userRow2['description']; ?>
It seems that the PHP variable is incompatible with html link :(
so I want to know what is the proper method.
TIA...
echo those variables there like the following.
<?php echo $userRow2['description']; ?>
Please use a template engine for these kinds of things...
Use one of:
smarty
twig
mustache
php-view
These will brighten up your day and remove the complexity out of your html files
You can also pass all your GET params in an associative array, and use:
http_build_query($params)
so:
or in your way:
<?php echo $userRow2['description']; ?>
You can also build html/php mix with heredoc:
http://www.hackingwithphp.com/2/6/3/heredoc
it seems that the php variable is incompatible with html link
Well, PHP runs server-side. HTML is client-side. So there's no way for client-side code to interpret PHP variables.
You need to enclose server-side code in <?php ?> tags in order for it to execute on the server (like you already do elsewhere). Otherwise the server just treats it as any other HTML and returns it to the browser. Something like this:
<?php echo $userRow2['description']; ?>
As you can see, that gets a bit messy. But you can put the whole thing in one echo statement:
echo "$userRow2[description]";
Notice how the double-quotes needed to be escaped in that one, but since the whole thing was a double-quoted string the variables contained therein would expand to their values.
There are readability pros and cons either way, so it's up to you how you want to present it.
you should use this
<?php echo $userRow2['description']; ?>
or
<?=$userRow2['description']?>
You can also use Here Doc Syntax
<?php
//test variables
$inst_id = 1;
$description = "Test 1";
$eof = <<<EOF
$description
EOF;
//test output
echo $eof;
http://php.net/manual/en/language.types.string.php#language.types.string.syntax.heredoc
I'm quite new here. I'm trying to make a blog/journal site that allows users to post their own journal. I'm still quite reluctant on making it because I am really afraid of malicious code injections.
So here's a sample code:
<?php
$test = "<b>blah</b>"; //User input from SQL
echo "$test";
?>
What will come out is just the word "blah" in bold right? What I was trying to achieve was to echo "<b>blah</b>" instead. I don't want people to put some PHP codes that can actually mess up my whole web page. Please keep in mind that the variable $test is actually a MYSQL query, so that variable will be needed as an example. I know you can do echo '$test'; but it just comes out as "$test" instead. I feel like pulling my hair out I can't figure it out yet.
The second solution I know of is the htmlspecialchars(); function, but I want the strings to display as what I typed, not the converted ones...
Is there any way I can do that?
I think the OP wants the HTML itself to be output to the page, and not have the tags stripped. To achieve this, you can run the string first through htmlentities()
$test = '<b>blah</b>';
echo htmlentities($test);
This will output:
<b>blah</b>
Which will render in the page as
<b>blah</b>
Echo don't execute PHP code from string. This is impossible and this is not security hole in your code.
You can use a template engine like Twig for exemple.
If htmlspecialchars(); is not the one you are looking for, try the header() option.
header('Content-type: text/plain');
When you are gonna give <b>Hi</b> to a browser, it will be displayed in Bold and not the text be returned. But you can try this way, outputting it inside a <textarea></textarea>.
Or the other way is to use htmlentities():
<?php
$test = "<b>blah</b>"; //User input from SQL
echo htmlentities("$test");
?>
I'm trying to make a dynamic menu in my web, in which only some pages from each section will appear.
The code I wrote was:
$menulist=array();
$menulist[1]='file1%#16';
$menulist[2]='file2%#9';
$menulist[3]='file3%#19';
$menulist[4]='file4%#8';
$menulist[5]='file5%#13';
$menulist[6]='file6%#14';
$menulist[7]='file7%#10';
$menulist[8]='file8%#23';
$menulist[9]='file9%#19';
$menulist[10]='file10%#18';
$menulist[11]='file11%#12';
function actualizaciones($matriz)
{
$linea=explode("%#",$matriz);
echo '<li><a href="first_chunk_of_URL'.$linea[0].'middle_chunk_of_url'.$linea[1].'last_chunk_of_URL">'.${$linea[0]}[$linea[1]].'</li>;
}
echo '<ul>';
array_walk($menulist,'actualizaciones');
echo '</ul>';
Every $linea[0] string is the name of another array (not shown in this code) which contains the text that should be in every possible link corresponding to every key passed by $linea[1].
I must have done something wrong, because the hyperlinks work fine but there's no text showing on them.
use the simple character like below
echo '<li><a href="first_chunk_of_URL'.$linea[0].'middle_chunk_of_url'.$linea[1].'last_chunk_of_URL">'.${$linea[0]}[$linea[1]].'<li>';
and the problem in your code is
.'</li>;
^^^^^
here is the problem it should be
.'</li>';
If I'm reading the question right, you're asking how to use variable variables in PHP.
This can be done using the double-dollar syntax - ie $$linea[0]. See the PHP manual for more info: http://uk.php.net/manual/en/language.variables.variable.php
But if that is what you're doing, I would say you're not writing good code: if variable variables are involved, there's almost always a better way of doing it.
Can't really offer much better assistance here without understanding more about what you're trying to do, but it sounds like you should be using subarrays rather than separate named variables for everything.
Hope that helps.
I'm using a while statement on this and I can echo each row fine i.e
echo $row['myrow'];
but what I want is to have the result put into a link like so:
echo "<img src='http://www.mysite.com/images/$row['myrow'].jpg'>";
But it doesn't work. What am I doing wrong?
Either echo it this way:
echo "<img src='http://www.mysite.com/images/{$row['myrow']}.jpg'>";
Or, IMHO much better, this way:
echo "<img src='http://www.mysite.com/images/".$row['myrow'].".jpg'>";
Give the documentation on double quoted-strings a quick refresh.
Another nice way to do it is to only use PHP for the dynamic part of the code. I think it results in nicer looking code.
<img src="http://www.mysite.com/images/<?php echo $row['myrow']; ?>.jpg">
Then of course the whole img tag should not be in a PHP code block, since it regular HTML.
You need to take care of your quotes...
Try this:
echo '<img src="http://www.mysite.com/images/'.$row['myrow'].'.jpg" />';
Also notice that you didn't close the element.
Accessing array elements and object properties/methods inside the string must be enclosed in curly braces (string parsing)
echo "<img src='http://www.mysite.com/images/{$row['myrow']}.jpg'>";
to make it complete, lol
echo "<img src='http://www.mysite.com/images/$row[myrow].jpg'>";
you have write your site url on place of example.com
echo "<img src='example.com/images/'.$row['myrow'].'.jpg'>";
I barely know how to use PHP and I can't seem to make my code show an image if a condition proves true. This is the code:
<?php
$search=get_search_query();
$first=$search[0];
if ($first=="#"){
}
?>
I tried writing this thinking it would work and it didn't:
echo "<html>";
echo "<img src='http://chusmix.com/Imagenes/grupos/lujan.jpg'>";
Also I tried a code I found which started with the function: header() but it caused a tremendously long error, which said something like header already defined.
Thanks
You have used 'double quotes' incorrectly in the echo statement.
Try the following:
echo "<img src='http://chusmix.com/Imagenes/grupos/lujan.jpg' alt='Preview not available' />"
Regards,
Mahendra Liya.
You should var_dump($first) to know what it contains
check if the condition is really getting true
and also put single quote inside the double quote.
if ($first=="#"){
echo 'yes it is true';
echo "<img src='http://chusmix.com/Imagenes/grupos/lujan.jpg'>";
}
close the img tag
The part of the query string starting with # (so-called "hash") is not being sent to the server. That is, if your page is called like myblog.com/foo?bar=baz#quux, you php script will only receive myblog.com/foo?bar=baz. You need javascript if you want to handle urls with hashes.