如何将其转换为php字符串并将其回显到<;选择>;?
这让我发疯,我不能让它工作 这就是我所做的:如何将其转换为php字符串并将其回显到<;选择>;?,php,html,Php,Html,这让我发疯,我不能让它工作 这就是我所做的: $v = "<?php if($test== 'lala' || $test== 'kaka'){echo selected='selected';}?>"; $b = "<?php if($test== 'tada' || $test== 'sada'){echo selected='selected';}?>"; <select> <option <?php echo $v; ?>
$v = "<?php if($test== 'lala' || $test== 'kaka'){echo selected='selected';}?>";
$b = "<?php if($test== 'tada' || $test== 'sada'){echo selected='selected';}?>";
<select>
<option <?php echo $v; ?> value="lala">lala</option>
<option <?php echo $b; ?> value="tada">tada</option>
</select>
$v=”“;
$b=“”;
value=“tada”>tada
在下拉列表中,我看到:value=“tada”>tada,应该是tada您的报价有误 更改:
$v = "<?php if($test== 'lala' || $test== 'kaka'){echo selected='selected';}?>";
$b = "<?php if($test== 'tada' || $test== 'sada'){echo selected='selected';}?>";
$v=”“;
$b=“”;
致:
value=“tada”>tada
它不会被解释,只需正常分配变量,然后回显结果:$v=($test='lala'| |$okkon=='kaka')?'选定':''代码>三元示例返回到基本php语法,然后在理解它之后开始想象。
<option <?php if($test== 'lala' || $test== 'kaka'){echo 'selected=selected';}?> value="lala">lala</option>
<option <?php if($test== 'tada' || $test== 'sada'){echo 'selected=selected';} ?> value="tada">tada</option>