Adding an HTML link with PHP

Question

i have this problem i hope some of you can help me with. I try to make a html link that get the link from my database.

echo "<p><pre><a href=$Feed['socialfacebook']><img src=facebook-24.png></a></pre></p>

It looks like you're missing the double quote at the end. – mark Oct 23 '14 at 15:23 — mark, Oct 23 '14 at 15:23

score 2 · Answer 1 · answered Oct 23 '14 at 15:21

2

You need to put the link in '".$variable ."' and the image src should be in ' ', too.

echo "<p><pre><a href='".$Feed['socialfacebook']."'><img src='facebook-24.png'></a></pre></p>";

answered Oct 23 '14 at 15:21

baao

Just copy the entire code, works exactly like it should – baao Oct 23 '14 at 15:25
Don't forget the closing "; at the end, you did in your question – baao Oct 23 '14 at 15:28

George · Answer 2 · 2014-10-23T15:29:04.477

1

Close your string with a double quote and end your statement with a semi-colon,
When getting array elements inside a string literal like you're doing, leave out the key quotes,
Use single quotes around your attribute in case of any spaces:

echo "<p><pre><a href='$Feed[socialfacebook]'><img src=facebook-24.png></a>";

edited Oct 23 '14 at 15:29

answered Oct 23 '14 at 15:22

George

Do you know any difference between `""`, `""` and `""` besides readability (performance, compatibility version, ...) ? – Brewal Oct 23 '14 at 15:34
@Brewal See this question: http://stackoverflow.com/questions/16790501/php-variable-interpolation-vs-concatenation – George Oct 23 '14 at 15:35

vpzomtrrfrt · Accepted Answer · 2014-10-24T01:46:30.970

1

Double-quoted strings don't work with array elements quite like that. You should close and concatenate the strings.

echo "<p><pre><a href=".$Feed['socialfacebook']."><img src=facebook-24.png></a></pre></p>"

You might also want to put some quotes on your attributes.

edited Oct 24 '14 at 01:46

answered Oct 23 '14 at 15:26

vpzomtrrfrt

This worked perfectly! i will accept it in 6 minutes when stackoverflow allows me too! thank you so much! and thanks to all of you guys! – Anoxy Oct 23 '14 at 15:29
1

This answer actually missed the quotes for the html attributes – Brewal Oct 23 '14 at 15:36

score 0 · Answer 4 · answered Oct 23 '14 at 15:37

0

u need to put the variable $Feed['socialfacebook'] in " . $Feed['socialfacebook'] ."

echo "<p><pre><a href=" . $Feed['socialfacebook'] . "<img src=facebook-24.png></a></pre></p>";

and don't forget the ;

answered Oct 23 '14 at 15:37

Yassine

azhpo · Answer 5 · 2014-10-23T15:33:16.360

-1

Shouldn't it be:

echo "<p><pre><a href='$Feed["socialfacebook"]'><img src=facebook-24.png></a>";

Don't forget that is href="".

edited Oct 23 '14 at 15:33

answered Oct 23 '14 at 15:23

azhpo

5 Answers5