Gradient over img tag using css

Question

I want to place a gradient over an tag. src attribute of the tag is angular-item. For example:

Answer 1

For 2020, mask-image can work well. It works in modern browsers (not IE, -webkit- prefix in many browsers currently). https://caniuse.com/#feat=css-masks

img {
   height: 200px;
   width: auto;
   mask-image: linear-gradient(to bottom, rgba(0,0,0,0) 0%,rgba(0,0,0,0.65) 100%);
   -webkit-mask-image: linear-gradient(to bottom, rgba(0,0,0,0) 0%,rgba(0,0,0,0.65) 100%);
}

   <img src="http://i.imgur.com/HDssntn.jpg" />

Answer 2

try placing a div over the image in question and placing the gradient on the div instead of the image.

Answer 3

I recommend you to set background-color:black; to your container and then set class img{opacity:0.4}. Then you will get the same effect as you got with

backgroundImage:linear-gradient(rgba(0, 0, 0, 0.8),rgba(0, 0, 0, 0.8),rgba(0, 0, 0, 0.8),rgba(0, 0, 0, 0.8)),url(img_url))

My example on Slide:

.Slide {
    position: relative;
    border: 1px solid blue;
    min-width: 100%;
    height: 100%;
    transition: 0.5s;
    background-color: rgb(0, 0, 0);
}

.Slide img{
    position: relative;
    border: 1px solid blue;
    min-width: 100%;
    height: 100%;
    transition: 0.5s;
    opacity: 0.4;
}

Answer 4

With z-index :

You may use a container and put the gradient on that container. Then use a negative z-index to position image behind the gradient.

.pickgradient {
  display:inline-block;
  background: -moz-linear-gradient(top, rgba(0,0,0,0) 0%, rgba(0,0,0,0.65) 100%); /* FF3.6+ */
  background: -webkit-gradient(linear, left top, left bottom, color-stop(0%,rgba(0,0,0,0.65)), color-stop(100%,rgba(0,0,0,0))); /* Chrome,Safari4+ */
  background: -webkit-linear-gradient(top, rgba(0,0,0,0) 0%,rgba(0,0,0,0.65) 100%); /* Chrome10+,Safari5.1+ */
  background: -o-linear-gradient(top, rgba(0,0,0,0) 0%,rgba(0,0,0,0.65) 100%); /* Opera 11.10+ */
  background: -ms-linear-gradient(top, rgba(0,0,0,0) 0%,rgba(0,0,0,0.65) 100%); /* IE10+ */
  background: linear-gradient(to bottom, rgba(0,0,0,0) 0%,rgba(0,0,0,0.65) 100%); /* W3C */
  filter: progid:DXImageTransform.Microsoft.gradient( startColorstr='#a6000000', endColorstr='#00000000',GradientType=0 ); /* IE6-9 */
}

img{
  position:relative;
  z-index:-1;
  display:block;
  height:200px; width:auto;
}

<div class="pickgradient">
  <img src="http://i.imgur.com/HDssntn.jpg" />
</div>

---

With a pseudo element :

As commented, you can also use a pseudo element with the gradient and absolute positioning to put the gradient over the image :

.pickgradient{
  position:relative;
  display:inline-block;
}
.pickgradient:after {
  content:'';
  position:absolute;
  left:0; top:0;
  width:100%; height:100%;
  display:inline-block;
  background: -moz-linear-gradient(top, rgba(0,0,0,0) 0%, rgba(0,0,0,0.65) 100%); /* FF3.6+ */
  background: -webkit-gradient(linear, left top, left bottom, color-stop(0%,rgba(0,0,0,0.65)), color-stop(100%,rgba(0,0,0,0))); /* Chrome,Safari4+ */
  background: -webkit-linear-gradient(top, rgba(0,0,0,0) 0%,rgba(0,0,0,0.65) 100%); /* Chrome10+,Safari5.1+ */
  background: -o-linear-gradient(top, rgba(0,0,0,0) 0%,rgba(0,0,0,0.65) 100%); /* Opera 11.10+ */
  background: -ms-linear-gradient(top, rgba(0,0,0,0) 0%,rgba(0,0,0,0.65) 100%); /* IE10+ */
  background: linear-gradient(to bottom, rgba(0,0,0,0) 0%,rgba(0,0,0,0.65) 100%); /* W3C */
  filter: progid:DXImageTransform.Microsoft.gradient( startColorstr='#a6000000', endColorstr='#00000000',GradientType=0 ); /* IE6-9 */
}

img{
  display:block;
  height:200px;width:auto;
}

<div class="pickgradient">
  <img src="http://i.imgur.com/HDssntn.jpg" />
</div>