Multiply all td values(numbers) from a table with 3.8 - php

Question

My table is here http://jsfiddle.net/wqk7Lauz/
All I need is to have all values from td:nth-of-type(3) multiplied with 3.8 on page load.
HTML
$('td:nth-of-type(3)').text(parseFloat($('td:nth-of-type(3)').text()) * 3.8)
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table class="table table-bordered mydatatable table-hover display responsive nowrap">
<tbody>
<tr>
</tr>
<tr>
<th>ID</th>
<th>Nome</th>
<th>Preço por 1000</th>
<th>Quantidade mínima</th>
<th>Quantidade máxima</th>
<th>Descrição</th>
</tr>
<tr>
<td>9</td>
<td>Product </td>
<td>
R$1.86
</td>
<td>100</td>
<td>10000</td>
<td>S1 </td>
</tr>
<tr>
<td>10</td>
<td>Product </td>
<td>
R$3.98
</td>
<td>1000</td>
<td>100000</td>
<td>S1</td>
</tr>
</tbody>
</table>

Answer 1

You're very close, you just need to use the callback version of text so you're only handling the text of each individual element, and remove the R$ at the beginning:
$('td:nth-of-type(3)').text(function() {
return parseFloat($(this).text().replace("R$", "")) * 3.8;
});
Example:
$('td:nth-of-type(3)').text(function() {
return parseFloat($(this).text().replace("R$", "")) * 3.8;
});
<table class="table table-bordered mydatatable table-hover display responsive nowrap">
<tbody>
<tr>
</tr>
<tr>
<th>ID</th>
<th>Nome</th>
<th>Preço por 1000</th>
<th>Quantidade mínima</th>
<th>Quantidade máxima</th>
<th>Descrição</th>
</tr>
<tr>
<td>9</td>
<td>Product </td>
<td>
R$1.86
</td>
<td>100</td>
<td>10000</td>
<td>S1 </td>
</tr>
<tr>
<td>10</td>
<td>Product </td>
<td>
R$3.98
</td>
<td>1000</td>
<td>100000</td>
<td>S1</td>
</tr>
</tbody>
</table>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
If you want to add back the R$, and perhaps limit the result to a specific number of places, you can do that with string concatenation and toFixed:
$('td:nth-of-type(3)').text(function() {
return "R$" + (parseFloat($(this).text().replace("R$", "")) * 3.8).toFixed(2);
});
Example:
$('td:nth-of-type(3)').text(function() {
return "R$" + (parseFloat($(this).text().replace("R$", "")) * 3.8).toFixed(2);
});
<table class="table table-bordered mydatatable table-hover display responsive nowrap">
<tbody>
<tr>
</tr>
<tr>
<th>ID</th>
<th>Nome</th>
<th>Preço por 1000</th>
<th>Quantidade mínima</th>
<th>Quantidade máxima</th>
<th>Descrição</th>
</tr>
<tr>
<td>9</td>
<td>Product </td>
<td>
R$1.86
</td>
<td>100</td>
<td>10000</td>
<td>S1 </td>
</tr>
<tr>
<td>10</td>
<td>Product </td>
<td>
R$3.98
</td>
<td>1000</td>
<td>100000</td>
<td>S1</td>
</tr>
</tbody>
</table>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>

Answer 2

Hi change your script function with this
$(".table tr td:nth-child(3)").each(function(){ //Loop
//alert($(this).text())
var thrdtd = $(this).text(); //Getting value of td
onlyno = thrdtd.replace('R$', ''); // Removing R$
//akert(thrdtd);
$(this).text('R$ '+ parseFloat(onlyno * 3.8).toFixed(2)) // Adding R$ and also multiplying at the same time and update that result to td again
});
In this first i am getting value of every third td value through loop
$(".table tr td:nth-child(3)").each(function(){
and then removing character from it before multiplying and then after multiplying i am updating the same td value

Answer 3

You can use the below code:
$('td:nth-of-type(3)').text("R$" + (parseFloat($(this).text().replace("R$", "")) * 3.8).toFixed(2));