Php jQuery中JSON对象的问题

我无法从jquery中的json对象获取值。它在开发工具中抛出错误消息（“uncaughtsyntaxerror:unexpectedtokenh”）

如何从JSON对象中获取值

请让我知道

PHP代码：（未编辑）

几个打字错误-

---

1-

谢谢，它正在工作。但我需要在输出中显示双引号。
[
    {
        "qus_type": "choice",

        "qus_text": "Which one is if statement?",

        "correct_answer": "<?php$t=date("H");if ($t<"20") {  echo "Haveagoodday!";}?>",

        "choice1": "<?php$t=date("H");if ($t<"20") {  echo "Haveagoodday!";} else {  echo                "Haveagoodnight!";}?>",

        "choice2": "<?php$t=date("H");if ($t<"20") {  echo "Haveagoodday!";}?>",

        "choice3": "<?php$t=date("H");if ($t<"10") {  echo "Haveagoodmorning!";} elseif ($t<"20") {  echo "Haveagoodday!";} else {  echo "Haveagoodnight!";}?>",

        "choice4": "",

        "choice5": ""
    }
]

jsonObj = jQuery.parseJSON(ques_json);

window.jQuery(jsonObj).each(function(index, item) {
    console.log(item.qus_type);
    console.log(item.qus_text);
    console.log(item.correct_answer);
    console.log(item.choice1);
    console.log(item.choice2);
    console.log(item.choice3);
    console.log(item.choice4);
    console.log(item.choice5);

})

"choice3": "<?php$t=date('H');if ($t < 10) {  echo 'Haveagoodmorning!';} elseif ($t < 20) {  echo 'Haveagoodday!';} else {  echo 'Haveagoodnight!';}?>",