Gradient over img tag using css - html

Question

I want to place a gradient over an <img> tag. src attribute of the tag is angular-item. For example:
<img src={{value.angitem.image}}>
I've tried to make css class:
.pickgradient {
background: -webkit-gradient(linear, left top, left bottom, color-stop(0%,rgba(0,0,0,0)), color-stop(100%,rgba(0,0,0,0.65)));
}
and
<img src={{value.angitem.image}} class="pickgradient ">
but it doesn't work. What should I do?

Answer 1

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>

Answer 2

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 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

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