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Home/ Questions/Q 7678967
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Editorial Team
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Editorial Team
Asked: May 31, 20262026-05-31T17:46:05+00:00 2026-05-31T17:46:05+00:00

I have written a small plugin to change the text size but it doesn’t

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I have written a small plugin to change the text size but it doesn’t seem to work properly.
$(‘#mypara’).changeTextSize(’30px’); doesn’t set the size to 30px but instead sets to 10px

jquery.myPlugin.js

jQuery.fn.changeTextSize = function(size){
alert(size);    //shows 30px        

var config = {
    'size' : '10px'
};
if(size) {
    $.extend(config,size);
}

alert(config.size); //shows 10px

return $(this).each(function() {
    $(this).css('font-size',config.size);
});

};

HTML

<script type = "text/javascript" src="jquery-1.7.1.js"></script>
<script type = "text/javascript" src="jquery.myPlugin.js"></script>

<script>
$(document).ready(function() {
    $('#mypara').changeTextSize('30px');    
}); 
</script>

</head>
<body>
<p id="mypara">dsfdsfdsf</p>

</body>
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-31T17:46:06+00:00Added an answer on May 31, 2026 at 5:46 pm

    The extend function is used to merge objects. You are trying to merge a string, size, into an object literal, config. You would need to pass an object as the argument to your plugin:

    $('#mypara').changeTextSize({
        size: '30px'
    });
    

    Here’s a working example.

    Alternatively, if that’s the only argument and there’s not going to be more options, it may be easier to forget about using extend at all, and just do something like this:

    jQuery.fn.changeTextSize = function(size){ 
        return this.css('font-size', size || "10px");
    };
    

    Notice how I’ve removed the each, since there is no need for it (most jQuery methods apply to every element in the matched set by default), and you also don’t need to pass this into jQuery since it is already an instance of jQuery.

    Here’s a working example of that one.

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