Member Avatar for laura301019

$dbQuery="select id, name from artists where name like '$keyword%' order by name asc limit 15";
$dbResult=mysql_query($dbQuery);
echo mysql_num_rows($dbResult)."\n";
while ($dbRow=mysql_fetch_array($dbResult)) {

// for each artist, return the number of albums belonging to that artist
// use a separate query here to obtain the number
// return the number after the artist's name in the echo statement, separated with a '^' (1b)

echo $dbRow["id"]."^".$dbRow["name"]."\n";

Have this code and as you can see i have to write another dbquery that will display the number of albums an artist has. I also have to modify this piece of code


function showArtists() {
if (ajaxObject.readyState==4) {
var artistsArray=ajaxObject.responseText.spl…
var numArtists=artistsArray[0];
var htmlStr="";
for (var i=1; i<=numArtists; i++) {
var artistDetails=artistsArray.split("^")…
htmlStr+="<div onmouseover='javascript:suggestOver(this… ";
htmlStr+="onmouseout='javascript:sugge… ";
htmlStr+="onclick='javascript:setArtis… ";

// add the number of albums in brackets after the artist's name in the next line (1b)
htmlStr+="class='suggest_link'>" + artistDetails[1] + "</div> ";

Anyone know how i write this, ive tried a few things but when i try it on the webpage it stops the ajax auto fill when typing.

  • 3 Contributors
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  • Latest Post Latest Post by Biiim

Recommended Answers

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Member Avatar for rch1231

Hello,

What you are looking for is something like this:

select artist, count(artist)
from mytable
group by artist
Member Avatar for Biiim 
SELECT id, name, count(*) as `albums` FROM artists
LEFT JOIN albums ON artists.id = albums.artistid
WHERE artists.name like '$keyword%'
GROUP BY artists.id
ORDER BY artists.name ASC
LIMIT 15;

the ajax doesn't look like the complete code either so hard to advise on that, if it's stopping things that worked before there is likely a javascript error in the console.

If you change to the query above, making `albums` your albums table and with the correct id fields it looks like you want to change this line:

echo $dbRow["id"]."^".$dbRow["name"]."\n";

to this

echo $dbRow["id"]."^".$dbRow["name"]."^".$dbRow["albums"]."\n";

then in the javascript you access it with "artistDetails[2]"

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