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Jquery cada uno, ¿cómo no tomar URL rotas?

¿Cómo no tomar imágenes con url rota?

 jQuery.each(jQuery('img'), function(k, o) { imageStack.push(jQuery(o)); });
almost 3 years ago · Juan Pablo Isaza
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Solo envíe las imágenes cuyo evento de load se haya disparado a imageStack .

Para hacer esto, puede vincular un detector de eventos de load a cada imagen que empuja la imagen a imageStack .

 const imageStack = [] $('img').on('load', function() { imageStack.push($(this)); }) $(window).on('load', function(){ //wait for every image to load, so imageStack will contain every loaded image console.log(imageStack) })
 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <img src="https://www.gravatar.com/avatar/0fdacb141bca7fa57c392b5f03872176?s=48&d=identicon&r=PG&f=1"> <img src="brokenurl">

Alternativamente, puede esperar a que se cargue cada imagen escuchando el evento de load de la ventana , luego verifique la propiedad naturalWidth de la imagen:

 const imageStack = [] $(window).on('load', function() { $('img').each(function() { if (this.naturalWidth != 0) { imageStack.push($(this)); } }) console.log(imageStack) })
 <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script> <img src="https://www.gravatar.com/avatar/0fdacb141bca7fa57c392b5f03872176?s=48&d=identicon&r=PG&f=1"> <img src="brokenurl">

almost 3 years ago · Juan Pablo Isaza Report
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