<?php $myconn=mysql_connect("localhost","root",""); mysql_select_db("login_test",$myconn);?><html> <head> <title>Edit</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <form method="POST"> <table> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <?php $strSql="select * from table_test"; $result=mysql_query($strSql,$myconn); while($row=mysql_fetch_array($result)) { ?> <tr> <td><input type="text" name="objname" value="<?php echo $row['objname']?>" /></td> <td><input type="text" name="objinfo" value="<?php echo $row['objinfo']?>" /></td> </tr> <?php } ?> </table> </form> <br> <div style="display:inline;"> <input type="submit" class="button_black"/> <input type="button" class="button_black" value="返回" onclick="javascript:document.location.href='test.php'"/> <?php print_r($_POST); if(!empty($_POST)){ $objname=$_POST["objname"]; $objinfo=$_POST["objinfo"]; $strSql="update table_test set objname='$objname',objinfo='$objinfo' where objname='$objname' or objinfo='$objinfo'"; $result=mysql_query($strSql,$myconn); echo"修改成功！"; } ?> </body><?php mysql_close($myconn); ?></html> 新人刚学php没多久
以上代码是我脑补的
结果是post没有传入值，点击提交后text里填写的值自动变回修改前的值
求指导！
回复讨论(解决方案)
if(!$myconn){
echo "连接失败".mysql_error();
}
$result=mysql_query($sql,$myconn);
if(!$result){
die(mysql_error());
}
加多些mysql错误判断，这样代码不清楚哪里错了。
if(!$myconn){
echo "连接失败".mysql_error();
}
$result=mysql_query($sql,$myconn);
if(!$result){
die(mysql_error());
}
加多些mysql错误判断，这样代码不清楚哪里错了。
多谢指导 但是问题还是没解决
你的表单里并没有提交按钮，也没有看到用于提交的 js 代码
所以 $_POST 是不可能得到任何数据的
你的表单里并没有提交按钮，也没有看到用于提交的 js 代码
所以 $_POST 是不可能得到任何数据的
提交按钮一定要写在form里才能获得post吗？ 长姿势了
你的表单里并没有提交按钮，也没有看到用于提交的 js 代码
所以 $_POST 是不可能得到任何数据的
<?php $myconn=mysql_connect("localhost","root",""); mysql_select_db("login_test",$myconn); if(!$myconn){ echo "连接失败".mysql_error(); }?><html> <head> <title>Edit</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <form method="POST"> <table> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <?php $strSql="select * from table_test"; $result=mysql_query($strSql,$myconn); while($row=mysql_fetch_array($result)) { $objname1=$row['objname']; $objinfo1=$row['objinfo']; ?> <tr> <td><input type="text" name="objname" value="<?php echo $objname1?>" /></td> <td><input type="text" name="objinfo" value="<?php echo $objinfo1?>" /></td> </tr> <?php } print_r($_POST); if(!empty($_POST)){ $objname=$_POST["objname"]; $objinfo=$_POST["objinfo"]; $strSql="update table_test set objname='$objname',objinfo='$objinfo' where objname='$objname1' or objinfo='$objinfo1'"; $result=mysql_query($strSql,$myconn); if(!$result){ die(mysql_error()); } echo"修改成功！"; /*$strSql="select * from table_test"; $result=mysql_query($strSql,$myconn); while($row=mysql_fetch_array($result)) { $objname1=$row['objname']; $objinfo1=$row['objinfo']; ?> <tr> <td><input type="text" name="objname" value="<?php echo $objname1?>" /></td> <td><input type="text" name="objinfo" value="<?php echo $objinfo1?>" /></td> </tr> <?php }*/ } ?> </table> <br> <div style="display:inline;"> <input type="submit" class="button_black"/> <input type="button" class="button_black" value="返回" onclick="javascript:document.location.href='test.php'"/> </form> </body><?php mysql_close($myconn); ?></html> 修改之后发现编辑以后只修改了表单最后一行的数据
是因为提出数据进行修改时选择的是打印表单的最后一行数据吗？
难道要按照id提出数据进行修改吗？
你的表单里并没有提交按钮，也没有看到用于提交的 js 代码
所以 $_POST 是不可能得到任何数据的
<?php $myconn=mysql_connect("localhost","root",""); mysql_select_db("login_test",$myconn); if(!$myconn){ echo "连接失败".mysql_error(); }?><html> <head> <title>Edit</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <form method="POST"> <table> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <?php $strSql="select * from table_test"; $result=mysql_query($strSql,$myconn); while($row=mysql_fetch_array($result)) { $objname1=$row['objname']; $objinfo1=$row['objinfo']; ?> <tr> <td><input type="text" name="objname" value="<?php echo $objname1?>" /></td> <td><input type="text" name="objinfo" value="<?php echo $objinfo1?>" /></td> </tr> <?php } print_r($_POST); if(!empty($_POST)){ $objname=$_POST["objname"]; $objinfo=$_POST["objinfo"]; $strSql="update table_test set objname='$objname',objinfo='$objinfo' where objname='$objname1' or objinfo='$objinfo1'"; $result=mysql_query($strSql,$myconn); if(!$result){ die(mysql_error()); } echo"修改成功！"; /*$strSql="select * from table_test"; $result=mysql_query($strSql,$myconn); while($row=mysql_fetch_array($result)) { $objname1=$row['objname']; $objinfo1=$row['objinfo']; ?> <tr> <td><input type="text" name="objname" value="<?php echo $objname1?>" /></td> <td><input type="text" name="objinfo" value="<?php echo $objinfo1?>" /></td> </tr> <?php }*/ } ?> </table> <br> <div style="display:inline;"> <input type="submit" class="button_black"/> <input type="button" class="button_black" value="返回" onclick="javascript:document.location.href='test.php'"/> </form> </body><?php mysql_close($myconn); ?></html> 修改之后发现编辑以后只修改了表单最后一行的数据
是因为提出数据进行修改时选择的是打印表单的最后一行数据吗？
难道要按照id提出数据进行修改吗？
你那个表单
这里只显示最后一条数据啊，每次循环下一次就会把上一次的值给覆盖掉，所以只会修改掉最后一条数据
你的表单里并没有提交按钮，也没有看到用于提交的 js 代码
所以 $_POST 是不可能得到任何数据的
<?php $myconn=mysql_connect("localhost","root",""); mysql_select_db("login_test",$myconn); if(!$myconn){ echo "连接失败".mysql_error(); }?><html> <head> <title>Edit</title> <link href="css.css" rel="stylesheet" type="text/css" /> </head> <body> <form method="POST"> <table> <tr class="table_head"> <td>objname</td> <td>objinfo</td> </tr> <?php $strSql="select * from table_test"; $result=mysql_query($strSql,$myconn); while($row=mysql_fetch_array($result)) { $objname1=$row['objname']; $objinfo1=$row['objinfo']; ?> <tr> <td><input type="text" name="objname" value="<?php echo $objname1?>" /></td> <td><input type="text" name="objinfo" value="<?php echo $objinfo1?>" /></td> </tr> <?php } print_r($_POST); if(!empty($_POST)){ $objname=$_POST["objname"]; $objinfo=$_POST["objinfo"]; $strSql="update table_test set objname='$objname',objinfo='$objinfo' where objname='$objname1' or objinfo='$objinfo1'"; $result=mysql_query($strSql,$myconn); if(!$result){ die(mysql_error()); } echo"修改成功！"; /*$strSql="select * from table_test"; $result=mysql_query($strSql,$myconn); while($row=mysql_fetch_array($result)) { $objname1=$row['objname']; $objinfo1=$row['objinfo']; ?> <tr> <td><input type="text" name="objname" value="<?php echo $objname1?>" /></td> <td><input type="text" name="objinfo" value="<?php echo $objinfo1?>" /></td> </tr> <?php }*/ } ?> </table> <br> <div style="display:inline;"> <input type="submit" class="button_black"/> <input type="button" class="button_black" value="返回" onclick="javascript:document.location.href='test.php'"/> </form> </body><?php mysql_close($myconn); ?></html> 修改之后发现编辑以后只修改了表单最后一行的数据
是因为提出数据进行修改时选择的是打印表单的最后一行数据吗？
难道要按照id提出数据进行修改吗？
你那个表单
这里只显示最后一条数据啊，每次循环下一次就会把上一次的值给覆盖掉，所以只会修改掉最后一条数据
谢谢！
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