Question

这里我试图从mysql表中获取数据并在html中填充它。一切正常但我想通过IF条件回显一个html按钮。我不知道这是否可行..如果它有效......任何人都可以告诉我如何使它工作？我得到的错误是......内部服务器错误：（

注意：我已经对我遇到问题的确切位置进行了评估

<?php
if(mysql_num_rows($sql)){
    while($rs = mysql_fetch_object($sql))
    {
    ?>  
        <tr>
            <td align="center"><?php echo $rs->cnf_name ;?></td>
            <td align="center"><?php echo $rs->address;?></td>
            <td align="center"><?php echo $rs->added_on;?></td>
            <td align="center"><input type="button" class="btn" value="Edit" onClick="window.parent.editCnf('<?php echo $rs->cnf_name;?>','<?php echo $rs->username ;?>','<?php echo $rs->password;?>','<?php $rs->type;?>','<?php echo $rs->person_name;?>','<?php echo $rs->address;?>','<?php echo $rs->mobile_no?>','<?php echo $rs->email;?>','<?php echo $rs->country;?>','<?php echo $rs->city;?>','<?php echo $rs->state;?>','<?php echo $rs->area;?>')" ></td>




//PROBLEM


            <td align="center">

            <?php
            if($rs->cnf_status == 0)
            {
            echo "<input type="button" class="btn" id="status" value="Activate" onClick="window.parent.deleteCnf(<?php echo $rs->user_id;?>,<?php echo $rs->cnf_status; ?>);">";
            }
            else
            {
            echo "<input type="button" class="btn" id="status" value="Activate" onClick="window.parent.deleteCnf(<?php echo $rs->user_id;?>,<?php echo $rs->cnf_status; ?>);">";

            }
?>          




            </td>
        </tr>
<?php
    }
}
else
{
?>      <tr>
            <td colspan="">No data to display</td>
        </tr>
<?php
}
?>      
    </tbody>
</table>

Answer 1

问题来自你用echo的引号; 要么

echo 'htmlcode_with_doublequotes';

或

if($rs->cnf_status == 0)
{
?>
       <input type="button" class="btn" id="status" value="Activate" onClick="window.parent.deleteCnf("<?php echo $rs->user_id;?>","<?php echo $rs->cnf_status; ?>");">
<?php
}
else
{
?>
       <input type="button" class="btn" id="status" value="Activate" onClick="window.parent.deleteCnf("<?php echo $rs->user_id;?>","<?php echo $rs->cnf_status; ?>");">
<?php
}

Answer 2

这似乎是按钮html的回显引用的问题。

if($rs->cnf_status == 0)
        {
        echo '<input type="button" class="btn" id="status" value="Activate" onClick="window.parent.deleteCnf(<?php echo $rs->user_id;?>,<?php echo $rs->cnf_status; ?>);">';
        }
        else
        {
        echo '<input type="button" class="btn" id="status" value="Activate" onClick="window.parent.deleteCnf(<?php echo $rs->user_id;?>,<?php echo $rs->cnf_status; ?>);">';

        }

Answer 3

问题在于你的语法。你必须逃避“在你的回声中：

<?php
    if($rs->cnf_status == 0)
        {
        echo "<input type=\"button\" class=\"btn\" id=\"status\" value=\"Activate\" onClick=\"window.parent.deleteCnf({$rs->user_id}, {$rs->cnf_status});\">";
        }
    else
        {
        echo "<input type=\"button\" class=\"btn\" id=\"status\" value=\"Activate\" onClick=\"window.parent.deleteCnf({$rs->user_id}, {$rs->cnf_status});\">";
        }
?>

OR

<?php
    if($rs->cnf_status == 0)
        {
        echo '<input type="button" class="btn" id="status" value="Activate" onclick="window.parent.deleteCnf(' . $rs->user_id . ', ' . $rs->cnf_status . ');">';
        }
    else
        {
        echo '<input type="button" class="btn" id="status" value="Activate" onclick="window.parent.deleteCnf(' . $rs->user_id . ', ' . $rs->cnf_status . ');">';
        }
?>

Answer 4

试试这个：

<?php
if($rs->cnf_status == 0)
{
echo '<input type="button" class="btn" id="status" value="Activate" onClick="window.parent.deleteCnf('.$rs->user_id.','.$rs->cnf_status.');">';
}
else
{
echo '<input type="button" class="btn" id="status" value="Activate" onClick="window.parent.deleteCnf('.$rs->user_id.','.$rs->cnf_status.');">';

}
?>

HTML和PHP嵌入

4 个答案: