1
I have a function in JavaScript
that mounts an image gallery. It had this whole function in Javascript
and it worked perfectly, now I need to insert the PHP
so that the user can make changes to the images, and I am passing the parameter to this function through the result of the query
in PHP
. It should work this way: Gallery title above, and below 4 different images of the result of query
. Below another title, and other images. The problem is that PHP
is passing the same parameter to the function, making the function return: gallery title and below 4 identical images(the first value of the query
), Other gallery title and other identical images(the second value of query
). How can I get the gallery to upload 4 different images to a gallery, not 4 identical images to different galleries ?
My role in JS
is like this:
function galerias(query){
var galeria = new Array();
//aqui eu pego a div que armazena as galerias
var imgs = document.querySelector("#gallery");
var x = 1;
imgs.innerHTML += "<div class='row'>";
imgs.innerHTML += "<div class='eight columns'>";
imgs.innerHTML += "<h4>Galeria "+x+"</h4>"
//passo o valor da query para o array galeria
for(var i = 0; i <= 3; i++){
galeria[i] = query;
}
e insiro dentro da div
for(var i = 0; i < galeria.length; i++){
imgs.innerHTML += "<img src='img/"+query+"'class='imgs-galeria'>";
}
imgs.innerHTML += "</div>";
imgs.innerHTML += "</div>";
imgs.innerHTML += "<a class='row' href='pics.html?gal="+x+"'><div class='twelve columns link'><p>Veja mais</p></div>";
x++;
}
My function in php:
function consultarDados($query){
$dbResult = array();
$conexao = mysql_connect($this->host, $this->usuario, $this->senha);
mysql_select_db($this->banco, $conexao);
$rs = mysql_query($query, $conexao);
while($rows = mysql_fetch_assoc($rs)){
array_push($dbResult, $rows);
}
return $dbResult;
mysql_close($conexao);
}
I insert this php in window.onload
page:
<?php
foreach($galeria_1 as $result){
echo'galerias("'.$result['nome'].'");';
}
?>
The visual result is this:
and it should be this:
Yes, he has all the positions !
– anuseranother
What is the result of the $result variable ?
– Kayo Bruno
$result['name'] returns the values of the column of my database: foto1_1.jpg foto1_2.jpg foto1_3.jpg foto1_4.jpg, and only the $result returns Array Array Array... is part is perfect, apparently
– anuseranother
So that’s the error my friend, realize that there are no galleries function you get "query" which was to be an array and not a name of one of the images, because in the course of the function you do a for mounting the images passing "query" as the name of the image in "src" but this "query" is a single value that you passed in "echo'galerias("'.$result['name'].'");';" .
– Kayo Bruno
Actually gave yes ! But how could I pass the array as parameter to the function...?
– anuseranother
Creates a Json!!!
– Kayo Bruno
Thank you for the reply !
– anuseranother
You are welcome, we are here to help each other! Strong hug my friend.
– Kayo Bruno