php转义括号字符解析错误

Question

我想我必须逃避{因为我收到错误 所以这是

function aq_field_icon_picker($field_id, $block_id, $icon, $default = '') {
    $output = '<div class="icon-picker" data-pickerid="fa" data-iconsets='{"fa":"Pick FontAwesome"}'>'; 
    return $output;
}

我怎样逃避{}？

Answer 1

你需要逃避＆＃39; {}

周围的字符

function aq_field_icon_picker($field_id, $block_id, $icon, $default = '') {
    $output = '<div class="icon-picker" data-pickerid="fa" data-iconsets=\'{"fa":"Pick FontAwesome"}\'>'; 
    return $output;
}

http://php.net/manual/en/regexp.reference.escape.php

或更改＆＃39;到＆＃34;

function aq_field_icon_picker($field_id, $block_id, $icon, $default = '') {
    $output = '<div class="icon-picker" data-pickerid="fa" data-iconsets="{"fa":"Pick FontAwesome"}">'; 
    return $output;
}

Answer 2

如果您尝试使用双引号而不是

，则使用单引号结束字符串

$output = '<div class="icon-picker" data-pickerid="fa" data-iconsets="{"fa":"Pick FontAwesome"}"">';

Answer 3

的单引号

data-iconsets='{"fa"

结束字符串，PHP不知道以下是什么。