如果PHP echo代码返回一个值，则运行脚本。 没什么

Question

I have the following code which returns the result I require inside of a tab. But because the JavaScript is in the same file it shows a blank tab when there is no data to show. I remove the JavaScript and the tab disappears. How can I run the JavaScript only if data is present so the tab will disappear? Or can I call it from another file?

<?php echo $block->escapeHtml($block->getProduct()->getData($this->getCode()));
?>
<script type="text/JavaScript">
    var commareplace = document.querySelectorAll("div > #bikefitment");
    for (var i = 0; i < commareplace.length; i++) {
        commareplace[i].innerHTML = commareplace[i].innerHTML.replace(/,/g, "<br />");
    }
</script>

Answer 1

Since you have to hide the tab if escapeHtml() returns empty value. So you can create a condition block to add the script if escapeHtml() returns a non-empty value.

<?php $EH = $block->escapeHtml($block->getProduct()->getData($this->getCode()));
if ($EH) {

echo $EH;
?>
<script type="text/JavaScript">
    var commareplace = document.querySelectorAll("div > #bikefitment");
    for (var i = 0; i < commareplace.length; i++) {
        commareplace[i].innerHTML = commareplace[i].innerHTML.replace(/,/g, "<br />");
    }
</script>

<?php } ?>