我有一个PHP的回声，我想用一个变量width属性来避免需要一个if语句。 我试图用这个代码：
<?php echo variable; ?>
它没有工作。
这里是我的代码：
echo "<div class='progress-bar progress-bar-success' role='progressbar' aria-valuenow='25' aria-valuemin='0' aria-valuemax='100' style='width: (Variable)'>";
Answer 1:
这是否工作：
echo "<div class='progress-bar progress-bar-success' role='progressbar' aria-valuenow='25' aria-valuemin='0' aria-valuemax='100' style='width: ".$variable."'>";
Answer 2:
不回应的HTML只能随声附和变量：
<div class='progress-bar progress-bar-success' role='progressbar' aria-valuenow='25' aria-valuemin='0' aria-valuemax='100' style='width: <?php echo $variable?>px'>;
请注意，我也呼应了可变后添加像素。
Answer 3:
<?php $variable=100'; ?>
echo "<div class='progress-bar progress-bar-success' role='progressbar' aria-valuenow='25' aria-valuemin='0' aria-valuemax='100' style='width: ( ".$variable.")'>";
Answer 4:
如果你想成为更看中你可以检查变量不为空：
<?php if($variable != ""):?>
<!-- this will be the output if variable has some value -->
<div class="progress-bar progress-bar-success" role="progressbar" aria-valuenow="25" aria-valuemin="0" aria-valuemax="100" style="width:<?php echo $variable; ?>px">
<?php else: ?>
<!-- some default value -->
<div class="progress-bar progress-bar-success" role="progressbar" aria-valuenow="25" aria-valuemin="0" aria-valuemax="100" style="width:100px">
<?php endif;?>
Answer 5:
尝试：
$variable = "10px";
echo "<div class='progress-bar progress-bar-success' role='progressbar' aria-valuenow='25' aria-valuemin='0' aria-valuemax='100' style='width: $variable'>";
当您使用“-string工程
文章来源: php variable in style (width) attribute
