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Replace symbol with tag using jquery

开发者 https://www.devze.com 2022-12-15 11:13 出处:网络
I have specification for products in text doc that I need to transfer into HTML. Might sound a bit strange but but what I\'m trying to do is replace paragraph tags into two table columns, so this:

I have specification for products in text doc that I need to transfer into HTML.

Might sound a bit strange but but what I'm trying to do is replace paragraph tags into two table columns, so this:

<p>Load (kg):5,000, 10,000, 20,000, 30,000</p>
<p>Excitation  Voltage: 10vdc recommended, 20vdc maximum</p>

into this

 <tr>
    <td>Load (kg)</td>
    <td>5,000, 10,000, 20,000, 30,000</td>
  </tr>

 <tr>
    <td>Excitation  Voltage</td>
    <td>10vdc recommended, 20vdc maximum</td>
  </tr>

To replace paragraph tag I've used function bellow

$("p").each(function 开发者_如何学运维() {
      $(this).replaceWith("<td>" + $(this).text() + '</td>');
    });

So I have to hook up to ' : ' to close and close and open another table data (column)

Any help much appreciated

Thank you in advance

Dom


This is specific to your particular example and assumes well-conditioned data. If you can have multiple colons in the text, you'll need to adjust use the limit argument of split to limit it to 2 results.

$("p").each(function () {
  var text = $(this).text().split(':');
  $(this).replaceWith("<td>" + text[0] + '</td><td>' + text[1] + '</td>');
});


You can use split and get the index 0 value to be placed in the first td and index 1 value placed in the second td.

Something like this

$("p").each ( function() {
    var splitArr = $(this).text().split(':');
    var str = "<table><tr><td>";
    str+= splitArr[0] + "</td><td>";
    str += splitArr[1] + "</td>";
    str += "</tr></table>";

    $(this).replaceWith ( str );
});


The easiest way is probably to use regExp:

$('p').each(function() {
    $(this).replaceWith("<tr><td>"+$(this).text().replace(/\:\s?/, '</td><td>')+'</td></tr>');
})

Also supports multiple cells and stripping white-space after the colon.

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