求助如何点击按钮更新数据?
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"><html><head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8"><title>Insert title here</title><script language="javascript" type="text/javascript"> function showtd(num) { document.getElementById("bg"+num).style.display = "none"; document.getElementById("show"+num).style.display = "block"; } function hidetd(num) { document.getElementById("bg"+num).style.display = 'block'; document.getElementById("show"+num).style.display = 'none'; } function change() { var tds = document.getElementsByTagName("td"); var t1 = document.getElementById("a").innerHTML; var txt = document.createElement("input"); txt.type = "text"; txt.value = t1; txt.id = "sort_t"; tds[29].appendChild(txt); txt.select(); } function mouseup(){ if (document.getElementById("sort_t") && event.srcElement.id != "sort_t") { var obj = document.getElementById("sort_t").parentElement; var txtValue = document.getElementById("sort_t").value; obj.innerText = txtValue; } } </script></head><body><form><table border="1"> <?php $db = mysql_connect('localhost','root','root'); mysql_select_db('ec',$db); if (!$db) { die('Could not connect: ' . mysql_error()); } $result = mysql_query("select * from goods"); echo "<tr><th>GoodsID</th><th>BarCode</th> <th>GoodsName</th> <th>Category</th> <th>Specifications</th> <th>Manufacturers</th> <th>Numbers</th> <th>Instructions</th> <th>Pictures</th> <th>Update</th> <th>Delete</th></tr>"; $n = 0;while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['id'] . "</td>"; echo "<td>" . $row['barcode'] . "</td>"; echo "<td>" . $row['goods_name'] . "</td>"; echo "<td>" . $row['category'] . "</td>"; echo "<td>" . $row['specifications'] . "</td>"; echo "<td>" . $row['manufacturers'] . "</td>"; echo "<td>" . $row['number'] . "</td>"; echo "<td>" . $row['instruction'] . "</td>"; echo "<td>" . "<img src = '$row[picture_url]' style='width:80px;height=60px'/>" . "</td>"; echo "<td id='bg$n'>". "<input id='btnshow$n' type='button' value='Update' onclick='showtd($n)'/>". "</td>". "<td id='show$n' style='display:none'>". "<input id='btnclose$n' type='button' value='Save' onclick='hidetd($n)'/>". "</td>"; echo "<td>"."<a href='delete.php?id=$row[id]'>"."Delete"."</a>"."</td>"; echo "</tr>"; ++$n; } mysql_close($db); ?> </table></form> </body></html>
回复讨论(解决方案)
1.给你的save按钮一个onclick方法 点击后触发
2.用ajax 提交参数到后台,调用php程序
3.然后将接收内容提交到数据库
4.成功or失败返回到页面
你这里要用到ajax来提交数据
1.给你的save按钮一个onclick方法 点击后触发
2.用ajax 提交参数到后台,调用php程序
3.然后将接收内容提交到数据库
4.成功or失败返回到页面
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"><html><head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8"><title>Insert title here</title><script language="javascript" type="text/javascript"> function showtd(num) { document.getElementById("bg"+num).style.display = "none"; document.getElementById("show"+num).style.display = "block"; } function hidetd(num) { document.getElementById("bg"+num).style.display = 'block'; document.getElementById("show"+num).style.display = 'none'; } function change() { var tds = document.getElementsByTagName("td"); var t1 = document.getElementById("a").innerHTML; var txt = document.createElement("input"); txt.type = "text"; txt.value = t1; txt.id = "sort_t"; tds[30].appendChild(txt); txt.select(); } function mouseup(){ if (document.getElementById("sort_t") && event.srcElement.id != "sort_t") { var obj = document.getElementById("sort_t").parentElement; var txtValue = document.getElementById("sort_t").value; obj.innerText = txtValue; } } </script></head><body> <form> <table border="1"> <?php $db = mysql_connect('localhost','root','root'); mysql_select_db('ec',$db); if (!$db) { die('Could not connect: ' . mysql_error()); } $result = mysql_query("select * from goods"); echo "<tr> <th>GoodsID</th> <th>BarCode</th> <th>GoodsName</th> <th>Category</th> <th>Specifications</th> <th>Manufacturers</th> <th>Numbers</th> <th>Instructions</th> <th>Pictures</th> <th>Update</th> <th>Delete</th> </tr>"; $n = 0; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['id'] . "</td>"; echo "<td id='a'>" . $row['barcode'] . "</td>"; echo "<td>" . $row['goods_name'] . "</td>"; echo "<td>" . $row['category'] . "</td>"; echo "<td>" . $row['specifications'] . "</td>"; echo "<td>" . $row['manufacturers'] . "</td>"; echo "<td>" . $row['number'] . "</td>"; echo "<td>" . $row['instruction'] . "</td>"; echo "<td>" . "<img src = '$row[picture_url]' style='width:80px;height=60px'/>" . "</td>"; echo "<td id='bg$n'>". "<input id='btnshow$n' type='button' value='Update' onclick='showtd($n);change()' />". "</td>". "<td id='show$n' style='display:none'>". "<input id='btnclose$n' type='button' value='Save' onclick='hidetd($n);mouseup()'/>". "</td>"; echo "<td>"."<a href='delete.php?id=$row[id]'>"."Delete"."</a>"."</td>"; echo "</tr>"; ++$n; } mysql_close($db); ?> </table></form> </body></html>
我表上红色的部分是一个数组,我想让这个数组变成动态的,要不然所有按钮都只操作一条数据。有什么好办法没?
1.给你的save按钮一个onclick方法 点击后触发
2.用ajax 提交参数到后台,调用php程序
3.然后将接收内容提交到数据库
4.成功or失败返回到页面
不好意思是第24行,刚才没标上颜色。
要是我做的话就是都设为隐藏的
一个.show()
一个.hide()
当点击完一个操作就触发函数转换到另一个按钮
建议lz选择jq选择器都很强大
搭配ajax 足够完成大量input提交工作
你这里要用到ajax来提交数据
<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN" "http://www.w3.org/TR/html4/loose.dtd"><html><head><meta http-equiv="Content-Type" content="text/html; charset=UTF-8"><title>Insert title here</title><script language="javascript" type="text/javascript"> function showtd(num) { document.getElementById("bg"+num).style.display = "none"; document.getElementById("show"+num).style.display = "block"; } function hidetd(num) { document.getElementById("bg"+num).style.display = 'block'; document.getElementById("show"+num).style.display = 'none'; } function change() { var tds = document.getElementsByTagName("td"); var t1 = document.getElementById("a").innerHTML; var txt = document.createElement("input"); txt.type = "text"; txt.value = t1; txt.id = "sort_t"; tds[29].appendChild(txt); txt.select(); } function mouseup(){ if (document.getElementById("sort_t") && event.srcElement.id != "sort_t") { var obj = document.getElementById("sort_t").parentElement; var txtValue = document.getElementById("sort_t").value; obj.innerText = txtValue; } } </script></head><body><form><table border="1"> <?php $db = mysql_connect('localhost','root','root'); mysql_select_db('ec',$db); if (!$db) { die('Could not connect: ' . mysql_error()); } $result = mysql_query("select * from goods"); echo "<tr><th>GoodsID</th> <th>BarCode</th><th>GoodsName</th><th>Category</th><th>Specifications</th><th>Manufacturers</th> <th>Numbers</th><th>Instructions</th><th>Pictures</th><th>Update</th><th>Delete</th></tr>"; $n = 0; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['id'] . "</td>"; echo "<td id='a'>" . $row['barcode'] . "</td>"; echo "<td>" . $row['goods_name'] . "</td>"; echo "<td>" . $row['category'] . "</td>"; echo "<td>" . $row['specifications'] . "</td>"; echo "<td>" . $row['manufacturers'] . "</td>"; echo "<td>" . $row['number'] . "</td>"; echo "<td>" . $row['instruction'] . "</td>"; echo "<td>" . "<img src = '$row[picture_url]' style='width:80px;height=60px'/>" . "</td>"; echo "<td id='bg$n'>". "<input id='btnshow$n' type='button' value='Update' onclick='showtd($n);change()' />". "</td>". "<td id='show$n' style='display:none'>". "<input id='btnclose$n' type='button' value='Save' onclick='hidetd($n);mouseup()'/>". "</td>"; echo "<td>"."<a href='delete.php?id=$row[id]'>"."Delete"."</a>"."</td>"; echo "</tr>"; ++$n; } mysql_close($db); ?> </table></form></body></html>
第24行的那个数组能不能改成动态的,要不然所有按钮都只操作同一条数据,应该怎么改啊?
<html><head><script type="text/javascript" src="/jquery/jquery.js"> </script><script type="text/javascript"> $(document).ready(function(){ $(".btn1").click(function(){ $(".btn1").hide(); $(".btn2").show(); }); $(".btn2").click(function(){ $(".btn1").show(); $(".btn2").hide(); }); }); </script></head><body><button class="btn1" style="display:none">update</button><button class="btn2">save</button></body></html>
刚才描述的不是很清楚 写了个测试的代码 你可以参考
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<html> <head> <script type="text/javascript" src="/jquery/jquery.js"></script> <script type="text/javascript"> $(document).ready(function……
谢谢你,我没学过ajax和jq。
items = tds.join(',');
好固执的孩纸啊
function change(obj) { //可以传递个当前对象过来 var tds = document.getElementsByTagName("td"); var t1 = document.getElementById("a").innerHTML; var txt = document.createElement("input"); txt.type = "text"; txt.value = t1; txt.id = "sort_t"; obj.parentNode.childNodes[1].appendChild(txt); //添加至第二个td,可根据需要修改 txt.select(); }
调用时,
change(this)
还有你的id命名重复了,这样只能获取第一个