使用PHP变量作为参数的JavaScript函数

Question

我已经搜索过逃避JS函数内部的参数，我已经尝试了很多建议，但都没有。这是代码：

echo '<a class="btn btn-xs btn-danger" onclick="delete('.$fetch['bicc'].')"><span class="glyphicon glyphicon-remove"></span> Remove</a>'

我总是收到这个错误：

SyntaxError: identifier starts immediately after numeric literal
delete(123456789_7_ZY7)

Answer 1

你需要添加引号，你的param是一个字符串类型：

echo '<a class="btn btn-xs btn-danger" onclick="delete(\''.$fetch['bicc'].'\')"><span class="glyphicon glyphicon-remove"></span> Remove</a>'

Answer 2

您忘记了\"

onClick="delete(\"'.$fetch['bicc'].'\")"><span class="glyphicon glyphicon-remove"></span> Remove</a>

所以应该是这样的：

echo '<a class="btn btn-xs btn-danger" onclick="delete(\"'.$fetch['bicc'].'\")"><span class="glyphicon glyphicon-remove"></span> Remove</a>'

Answer 3

根据经验，如果您想在JavaScript代码中使用PHP变量，则需要将它们包装在json_encode中：

echo '<a class="btn btn-xs btn-danger" onclick="delete('.json_encode($fetch['bicc']).')"><span class="glyphicon glyphicon-remove"></span> Remove</a>'

json_encode能够获取大多数PHP变量（字符串，数组，关联数组）并将它们正确编码为JavaScript副本（例如，为字符串添加引号）。

Answer 4

您应该添加引号，因为参数是字符串

echo "<a class=\"btn btn-xs btn-danger\" onclick=\"delete(\"${fetch['bicc']}\")\"><span class=\"glyphicon glyphicon-remove\"></span> Remove</a>";

Answer 5

$ fetch ['bicc']有字符串类型值吗？

如果是的话：

echo '<a class="btn btn-xs btn-danger" onclick="delete(\"'.$fetch['bicc'].'\")"><span class="glyphicon glyphicon-remove"></span> Remove</a>'

Answer 6

$data = str_replace('_', ' ', $fetch['bicc']);
echo '<a class="btn btn-xs btn-danger" onclick="delete('.$data.')"><span class="glyphicon glyphicon-remove"></span> Remove</a>'