Script html php - php

Question

In the script below, I currently have a fixed value in the array of $out variable. Example: $out[1]. What I need is for this fixed value (in this case 1), to be replaced by the str value of the function.
str is javascript
$out[1] is php
Someone can help me?
<script>
function teste2(str) {
var xhttp;
if (str == "") {
document.getElementById("cli").innerHTML = "";
return;
}
xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
document.getElementById("resimg").innerHTML =
"<img src=" + "<?php echo ("img/base/".$out[1]);?>" + ">"
}
};
xhttp.open("GET", "clique.php?q="+str, true);
xhttp.send();
}
</script>

Answer 1

As mentioned, it doesn't make sense why you are mixing PHP in your JS. If you need data from the server side usable by your client side JS, there are multiple approaches.
The preferred approach is to make an API endpoint you can call and return JSON.
Another approach (not recommended) is to output a JSON object in a script tag on server render and then access it like:
<script>
let data = <?php echo(json_encode($myData)); ?>;
</script>
To answer your question though, you can set the value with JS since the variable you want is client side:
function teste2(str) {
var xhttp;
if (str == "") {
document.getElementById("cli").innerHTML = "";
return;
}
xhttp = new XMLHttpRequest();
xhttp.onreadystatechange = function() {
if (this.readyState == 4 && this.status == 200) {
let src = 'img/base/' + str;
document.getElementById("resimg").innerHTML = '<img src="' + src + '">';
}
};
xhttp.open("GET", "clique.php?q="+str, true);
xhttp.send();
}

Answer 2

Well, PHP scripts run on the server side, while Javascript ones run on the client's browser.
Therefore, the value of str will only be resolved at runtime, when the function is called on the user's browser. There is no way for your PHP script to know in advance what the value will be, as your PHP script runs before any JS does.
I am sorry, but I believe you should re-imagine your design here.
Edit: I assume your $out is a PHP variable, right?