我想在JavaScript代码中调用PHP函数来回显一个值:
<?php
function ex()
{
$num=10;
echo($num);
}
echo '<script type="text/javascript">
function sam_click(clicked)
{
var x="<?php ex(); ?>";
alert(x);
return false;
}
</script>'?>
然而,它不起作用。你能帮忙吗?
答案 0 :(得分:1)
试试这个:
<?php
function ex()
{
$num=10;
echo($num);
}
echo '<script type="text/javascript">
function sam_click(clicked)
{
var x="' . ex() . '>";
alert(x);
return false;
}
</script>'?>
答案 1 :(得分:1)
试试:
...
var x="' . ex() . '";
...
答案 2 :(得分:1)
尝试仅回显您需要的内容
<?php
function ex()
{
$num=10;
echo($num);
}
?>
<script type="text/javascript">
function sam_click(clicked)
{
var x="<?php ex(); ?>";
alert(x);
return false;
}
</script>
答案 3 :(得分:0)
function ex()
{
$num=10;
echo $num;
return $num; // if you want to get JS alert
}
echo '<script type="text/javascript">
function sam_click(clicked)
{
var x='.ex().';
alert(x);
return false;
}
sam_click(); // if you want to get JS alert
</script>';
答案 4 :(得分:0)
而不是回显$ num,将其作为结果返回,然后将其附加到脚本部分。
<?php
function ex()
{
$num=10;
return ($num);
}
echo '<script type="text/javascript">
function sam_click(clicked)
{
var x="' . ex() . '";
alert(x);
return false;
}
</script>'?>
答案 5 :(得分:0)
heredoc和nowdoc的另一个解决方案:
<?php
// helper to call functions as nowdoc
// example: {$fn( function() )}
function fn($functionCall)
{
return $functionCall;
}
$fn = 'fn';
function ex()
{
return 20;
}
$js = <<<EOT
<script type="text/javascript">
function sam_click(clicked)
{
var x="{$fn(ex())}";
alert(x);
return false;
}
</script>
EOT;
echo $js;
详细了解PHPDoc: