I am trying to swap the form values in row1 with the form values of row2 without swapping the rows. Can someone show me away to achieve this in pure Javascript, vanilla JS, or jQuery. I made the table rows shorter with just two rows, but the actual table consists of 17 rows. Please look very closely at the ids and form values in the third example.
When the UP or DOWN button is not click, the table looks like this in simple form:
<form id="menuitems">
<table class="toolbaritems">
<tbody class="sortable">
<tr id="row1">
<td><button class="up_arrow">UP</button></td>
<td><input value="1></td>
<td><select><option="1" selected></option></select></td>
<td><select><option="1a" selected></option></select></td>
<td><img id="img1"></td>
</tr>
<tr id="row2">
<td><button class="down_arrow">DOWN</button></td>
<td><input value="2"></td>
<td><select><option="2" selected></option></select></td>
<td><select><option="2a" selected></option></select></td>
<td><img id="img2"></td>
</tr>
<tr><td><input type="submit" value="SAVE"></td></tr>
</tbody>
</table>
</form>
This the code currently - When the UP or DOWN buttons are clicked the table looks like this:
<form id="menuitems">
<table class="toolbaritems">
<tbody class="sortable">
<tr id="row2">
<td><button class="up_arrow">UP</button></td>
<td><input value="2"></td>
<td><select><option="2" selected></option></select></td>
<td><select><option="2a" selected></option></select></td>
<td><img id="img2"></td>
</tr>
<tr id="row1">
<td><button class="down_arrow">DOWN</button></td>
<td><input value="1"></td>
<td><select><option="1" selected></option></select></td>
<td><select><option="1a" selected></option></select></td>
<td><img id="img1"></td>
</tr>
<tr><td><input type="submit" value="SAVE"></td></tr>
</tbody>
</table>
</form>
This is what I am trying to accomplish - The values of the inputs should swap except for the tr. Notices the tr ids remain the same but form values are swapped:
Is there a way to achieve this in pure javascript, vanilla JS, or jquery. It will even be even better if this can be done with .html() instead of .val()
<form id="menuitems">
<table class="toolbaritems">
<tbody class="sortable">
<tr id="row1">
<td><button class="down_arrow">DOWN</button></td>
<td><input value="2"></td>
<td><select><option="2" selected></option></select></td>
<td><select><option="2a" selected></option></select></td>
<td><img id="img2"></td>
</tr>
<tr id="row2">
<td><button class="up_arrow">UP</button></td>
<td><input value="1"></td>
<td><select><option="1" selected></option></select></td>
<td><select><option="1a" selected></option></select></td>
<td><img id="img1"></td>
</tr>
<tr><td><input type="submit" value="SAVE"></td></tr>
</tbody>
</table>
</form>
Sorry it took awhile to get back to this. I am submitting this as another answer so that you can compare the two solutions. They are quite similar, and you may find it useful to compare the changes.
var tbl = $('table'),new_ndx,topRow,trStuff,botRow,brStuff;
$(document).on('click','button',function() {
var dir = $(this).attr('class');
var row = $(this).closest("tr");
var ndx = row.index();
//row.remove();
if (dir=='up_arrow'){
new_ndx = ndx-1;
topRow = tbl.find('tr').eq(new_ndx);
trStuff = topRow.html();
botRow = tbl.find('tr').eq(ndx);
brStuff = botRow.html();
topRow.html(brStuff);
botRow.html(trStuff);
} else {
new_ndx = ndx++;
topRow = tbl.find('tr').eq(new_ndx);
trStuff = topRow.html();
botRow = tbl.find('tr').eq(ndx);
brStuff = botRow.html();
topRow.html(brStuff);
botRow.html(trStuff);
}
});