我正在插入多個數據。當我從下拉列表中選擇一個項目時,會發生什么。它會保存所有選擇的數據。請先檢查下面的圖片 function sample($con){ $select = "SELECT * FROM project_tbl"; $select_result = mysqli_query($con,$select); if (mysqli_num_rows($select_result)> 0) { while ($row = mysqli_fetch_assoc($select_result)) { echo "<option value=".$row['project_name'].">" .$row['project_name']."</option>"; } } }* html代碼(我使用下拉列表填充數據庫中的數據)<td class="pro"> <select class="pro"> <option value=""><?php sample($con); ?></option> </select> </td>* ajax代碼(添加,保存和刪除)$(document).ready(function(){ var count = 1; $('#add').click(function(){ count = count + 1; var html_code = "<tr id='row"+count+"'>"; html_code += "<td class='pro'><select class='pro'><option value=''><?php sample($con); ?></option></select></td>"; html_code += "<td contenteditable='true' class='desc'></td>"; html_code += "<td contenteditable='true' class='comp'></td>"; html_code += "<td contenteditable='true' class='entry'></td>"; html_code += "<td contenteditable='true' class='remarks'></td>"; html_code += "<td><button type='button' name='remove' data-row='row"+count+"' class='btn btn-outline-danger btn-xs remove' title='remove'><i class='fa fa-times' aria-hidden='true'></i></button></td>"; html_code += "</tr>"; $('#crud_table').append(html_code); }); $(document).on('click', '.remove', function(){ var delete_row = $(this).data("row"); $('#' + delete_row).remove(); }); $('#save').click(function(){ var desc = []; var pro = []; var comp = []; var entry = []; var remarks = []; $('.desc').each(function(){ desc.push($(this).text()); }); $('.pro').each(function(){ pro.push($(this).text()); }); $('.comp').each(function(){ comp.push($(this).text()); }); $('.entry').each(function(){ entry.push($(this).text()); }); $('.remarks').each(function(){ remarks.push($(this).text()); });這是我保存時發生的情況。
我無法保存特定的選定值。它保存了數據庫中的所有數據
一只名叫tom的貓
2021-04-16 18:19:57