以PHP變量為參數的JavaScript函數

Question

我搜索了有關在JS函數中轉義參數的信息，並且嘗試了很多建議，但均無濟於事。 這是代碼：

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

我總是收到此錯誤：

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

Answer 1

您需要添加引號，您的param是字符串類型：

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

Answer 2

您忘記了\\"

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

所以應該像這樣：

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

Answer 3

根據經驗，如果要在JavaScript代碼中使用PHP變量，則需要將它們包裝在json_encode ：

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

json_encode能夠接受大多數PHP變量（字符串，數組，關聯數組），並將其正確編碼為與JavaScript對應的變量（例如，為字符串添加引號）。

Answer 4

您應該添加引號，因為參數是字符串

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

Answer 5

$ fetch ['bicc']是否具有字符串類型值？

如是：

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>'