要在javascript函数中将php变量作为参数传递吗?

if $id is not numeric you should write 

<?php
        while($gg=mysql_fetch_array($lg))
        {
    ?>
        <td id="add_td">
        <?php
        $id = $gg['p_id'];
        echo "<a onclick=cnf('Are you sure you want to delete that?','".$id."')>"; ?>Delete</a>
        </td>
<?php
        }
?>

检查语法

<?php
    while($gg=mysql_fetch_array($lg))
    {
?>
        <td id="add_td">
        <?php
        $id = $gg['p_id'];
        ?>
        <a onclick="cnf('Are you sure you want to delete that?',<?=$id?>);">Delete</a>
        </td>
<?php
     }
?>

唯一的ID在标签上也是必需的,如果在函数中传递ID而不是我建议的ID,则需要引用它,并给链接ID

function cnf(link,id) {
  if (confirm("Are you sure you want to delete "+id) {
    link.href = "p_update.php?d=" + id;          
    return true;
  }
  return false;
}

<?php
  while($gg=mysql_fetch_array($lg)) { 
    $id = $gg['p_id'];
?>
  <td id="add_td<?php echo $id; ?>"><a target="_parent" href="#" id="<?php echo $id; ?>" onclick="return cnf(this)">Delete</a></td>
<?php } ?>

foreach($mas as $k=>$v) {
  //echo $k.' = '.$v.'<br>';
  echo("

    <input type=\"button\" id=\"delete_id".$k."\" value=\"Blablabla\" onclick=\"alert('".$v."');\"/>

  ");
}

如果我把$k(index)放在那里,它会工作,但是如果string(value是string),我会得到一个错误

在HTML标记中,这可以工作,但不能作为函数参数!

echo "<a onclick=cnf('Are you sure you want to delete that?','$id')>";

另一个例子, $p 是一些php变量,

echo "<input type=button value='update' onclick=myfunc('$p')>"

如果$id是一个字符串文字,则在将其作为参数传递给<a中的cnf()函数时,应将其放入引号中。。。>标记:

echo "<a onclick=cnf('Are you sure you want to delete that?', '$id')>";