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menuOrder is an empty div

$("#menuOrder").append('<table>');

$(menus).each(function(i, menu) {
  $("#menuOrder").append('<tr><td>');
  $("#menuOrder").append(menu.text);
  $("#menuOrder").append('</td><td>');

  if (i > 0) {
    $("#menuOrder").append('<img src="/images/_images/up.png" />');
  } else {
    $("#menuOrder").append('<img src="/images/_images/blank.png" />');
  }

  if (i < menus.length - 1) {
    $("#menuOrder").append('<img src="/images/_images/down.png" />');
  }

  $("#menuOrder").append('</td>');
  $("#menuOrder").append('</tr>');
});

$("#menuOrder").append('</table>');

this code not work properly, how can I change it with minimum iterations?

menuOrder is an empty div

$("#menuOrder").append('<table>');

$(menus).each(function(i, menu) {
  $("#menuOrder").append('<tr><td>');
  $("#menuOrder").append(menu.text);
  $("#menuOrder").append('</td><td>');

  if (i > 0) {
    $("#menuOrder").append('<img src="/images/_images/up.png" />');
  } else {
    $("#menuOrder").append('<img src="/images/_images/blank.png" />');
  }

  if (i < menus.length - 1) {
    $("#menuOrder").append('<img src="/images/_images/down.png" />');
  }

  $("#menuOrder").append('</td>');
  $("#menuOrder").append('</tr>');
});

$("#menuOrder").append('</table>');

this code not work properly, how can I change it with minimum iterations?

• javascript
• jquery
• html-table

Share Improve this question Follow edited May 28, 2017 at 15:15 Brian Tompsett - 汤莱恩 5,8937272 gold badges6161 silver badges133133 bronze badges asked Apr 26, 2010 at 21:10 kusanagikusanagi 14.6k2222 gold badges8989 silver badges112112 bronze badges 1

• 2 It is remend to avoid lots of .append like this, instead when possible, build the string with HTML and then append – Fabiano Soriani Commented Apr 26, 2010 at 21:12

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 4

Try doing something like this:

var rows = [];
$(menus).each(function(i, menu) {
 var row = '<tr><td>'.menu.text.'</td><td>';

    if (i > 0) {
        row +='<img src="/images/_images/up.png" />';
    }
    else {
        row +='<img src="/images/_images/blank.png" />';
    }

    if (i < menus.length - 1) {
       row +='<img src="/images/_images/down.png" />';
    }

    row += '</td></tr>';
    rows.push(row);
});
$('<table></table>').append(rows.join('')).appendTo('#menuOrder');

This fills the rows array with a string that contains the HTML for each row. When its done creating all the rows then it creates a jQuery Object of a table element and appends all the rows to it. It then appends the table to #menuOrder.

Note that with append() you can't do this:

$("#menuOrder").append('</table>');

Instead of glueing together pieces of HTML strings, you must think in terms of plete DOM nodes. Examples of valid arguments for append():

.append($('.something'))
.append(document.createElement('div'))
.append('<div />') // automatically creates empty div element
.append('<tr><td></td></tr>') // automatically creates tr with td inside

menuOrder is an empty div

$("#menuOrder").append('<table>');

$(menus).each(function(i, menu) {
  $("#menuOrder").append('<tr><td>');
  $("#menuOrder").append(menu.text);
  $("#menuOrder").append('</td><td>');

  if (i > 0) {
    $("#menuOrder").append('<img src="/images/_images/up.png" />');
  } else {
    $("#menuOrder").append('<img src="/images/_images/blank.png" />');
  }

  if (i < menus.length - 1) {
    $("#menuOrder").append('<img src="/images/_images/down.png" />');
  }

  $("#menuOrder").append('</td>');
  $("#menuOrder").append('</tr>');
});

$("#menuOrder").append('</table>');

this code not work properly, how can I change it with minimum iterations?

menuOrder is an empty div

$("#menuOrder").append('<table>');

$(menus).each(function(i, menu) {
  $("#menuOrder").append('<tr><td>');
  $("#menuOrder").append(menu.text);
  $("#menuOrder").append('</td><td>');

  if (i > 0) {
    $("#menuOrder").append('<img src="/images/_images/up.png" />');
  } else {
    $("#menuOrder").append('<img src="/images/_images/blank.png" />');
  }

  if (i < menus.length - 1) {
    $("#menuOrder").append('<img src="/images/_images/down.png" />');
  }

  $("#menuOrder").append('</td>');
  $("#menuOrder").append('</tr>');
});

$("#menuOrder").append('</table>');

this code not work properly, how can I change it with minimum iterations?

• javascript
• jquery
• html-table

Share Improve this question Follow edited May 28, 2017 at 15:15 Brian Tompsett - 汤莱恩 5,8937272 gold badges6161 silver badges133133 bronze badges asked Apr 26, 2010 at 21:10 kusanagikusanagi 14.6k2222 gold badges8989 silver badges112112 bronze badges 1

• 2 It is remend to avoid lots of .append like this, instead when possible, build the string with HTML and then append – Fabiano Soriani Commented Apr 26, 2010 at 21:12

Add a ment  | 

2 Answers 2

Sorted by: Reset to default 4

Try doing something like this:

var rows = [];
$(menus).each(function(i, menu) {
 var row = '<tr><td>'.menu.text.'</td><td>';

    if (i > 0) {
        row +='<img src="/images/_images/up.png" />';
    }
    else {
        row +='<img src="/images/_images/blank.png" />';
    }

    if (i < menus.length - 1) {
       row +='<img src="/images/_images/down.png" />';
    }

    row += '</td></tr>';
    rows.push(row);
});
$('<table></table>').append(rows.join('')).appendTo('#menuOrder');

This fills the rows array with a string that contains the HTML for each row. When its done creating all the rows then it creates a jQuery Object of a table element and appends all the rows to it. It then appends the table to #menuOrder.

Note that with append() you can't do this:

$("#menuOrder").append('</table>');

Instead of glueing together pieces of HTML strings, you must think in terms of plete DOM nodes. Examples of valid arguments for append():

.append($('.something'))
.append(document.createElement('div'))
.append('<div />') // automatically creates empty div element
.append('<tr><td></td></tr>') // automatically creates tr with td inside

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