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>
我有一个错误,如:syntax error, unexpected T_CONSTANT_ENCAPSED_STRING, expecting ',' or ';'
in,但我不明白是什么问题,或者当我修改它时,鼠标在代码上不起作用。我哪里做错了?
鼠标移过和鼠标移出代码语法错误:
替换:
onmouseover="this.src=''/img/language_selection/us.png"''
加上下面的代码:
onmouseover="this.src=''/img/language_selection/us.png''"
^^^
单配额在双配额之外。所以每次mouseover和mouseout事件都要修改