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Home/ Questions/Q 9076391
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Editorial Team
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Editorial Team
Asked: June 16, 20262026-06-16T19:05:56+00:00 2026-06-16T19:05:56+00:00

The code below gives an undefined error on console for the variable mystyle .

  • 0

The code below gives an undefined error on console for the variable mystyle. I don’t get it as i defined the variable.

jQuery( '.styles_div' ).each( function() {
switch (styles) {
    case 'style1':
        var mystyle = $('#stylewrap').append('<div class="style1"></div> ');
        break;
    case 'style2':
        var mystyle = $('#stylewrap').append('<div class="style2"></div> ');
        break;
}
$("#search").autocomplete({
    delay: 0,
    minLength: 3,
    search: function( event, ui ) {mystyle.show();},
    ...
})
});
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-16T19:05:58+00:00Added an answer on June 16, 2026 at 7:05 pm

    mystyle needs to be defined before it is used – perhaps as a global var and not have the var keyword twice. It is also prudent to test the existence

    var mystyle;
    switch (styles) {
        case 'style1':
            mystyle = $('#stylewrap').append('<div class="style1"></div> ');
            break;
        case 'style2':
            mystyle = $('#stylewrap').append('<div class="style2"></div> ');
            break;
    }
    $("#search").autocomplete({
        delay: 0,
        minLength: 3,
        search: function( event, ui ) {if (mystyle) mystyle.show();},
        ...
    })
    

    if this is the complete code, then perhaps this code is simpler:

    var mystyle = $('#stylewrap').append('<div class="'+styles+'"></div> ');
    
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