Php从文件中读取错误

时间:2013-06-20 11:39:08

标签: php html

我有这个:

     echo '


     <div id="headlineatas">
     <br>
     <div id="adsground">
     <br><br><br>
     <p align="center"><span style="font-size: 44px;"> Language Selection </span></p>
<table height="400" width="1024">
        <tr>
        <td>


        <center><a href="index.php?lang=en" id="language1">
    <img title="English" src="/img/language_selection/us_first.png" onmouseover="this.src=\'/img/language_selection/us.png"\' onmouseout="this.src=\'/img/language_selection/us_first.png"\' />
</a>

            </center>

        </td>
        <td>

        <center><a href="index.php?lang=ro" id="language2">
    <img title="Romanian" src="/img/language_selection/ro_first.png" onmouseover="this.src=\'/img/language_selection/ro.png"\' onmouseout="this.src=\'/img/language_selection/ro_first.png"\' />
</a>


        </center>

        </td>
        </tr>
        </table>



</div></div></center>

我的错误如:syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';'但我不明白是什么问题,或者当我修改它时,鼠标悬停在代码上并不起作用。哪里出错?

1 个答案:

答案 0 :(得分:1)

鼠标悬停和鼠标输出代码中的语法错误:

替换:

onmouseover="this.src=\'/img/language_selection/us.png"\'

以下代码:

onmouseover="this.src=\'/img/language_selection/us.png\'"
                                                      ^^^

此处单一配额在双重配额之外。所以每次mouseover和mouseout事件都要改变它。