Php 使用AJAX选择2始终未找到结果

Php 使用AJAX选择2始终未找到结果,php,json,ajax,select,select2,Php,Json,Ajax,Select,Select2,在本教程之后,我将尝试使用ajax检索学生列表。我的意见是: <select class="form-control" id="student_id" name="student_id"></select> JSON响应类似于 [{"id":1,"text":"EJAJUL HAQUE"},{"id":2,"text":"MUSTAK AHMED"},{"id":3,"text":"AZAD HOQUE"},{"id":4,"text":"ANJARUL HAQUE"}

在本教程之后,我将尝试使用ajax检索学生列表。我的意见是:

<select class="form-control" id="student_id" name="student_id"></select>
JSON响应类似于

[{"id":1,"text":"EJAJUL HAQUE"},{"id":2,"text":"MUSTAK AHMED"},{"id":3,"text":"AZAD HOQUE"},{"id":4,"text":"ANJARUL HAQUE"},{"id":5,"text":"ARIFUL ISLAM"},{"id":6,"text":"SANJITA KHATUN"},{"id":7,"text":"MARINA SULTANA"},{"id":8,"text":"SULTANA BEGUM"},{"id":9,"text":"SABINA KHATUN"},{"id":10,"text":"JENIFA AHMED"},{"id":11,"text":"REJUON AHMED"},{"id":12,"text":"FARITA KHATUN"},{"id":13,"text":"HABIJUL HOQUE"},{"id":14,"text":"JENIFA MUSKAN"},{"id":15,"text":"JOTUFA WAHID"},{"id":16,"text":"MAUSUMI BEGUM"},{"id":17,"text":"ABJARUL RAHMAN"},{"id":18,"text":"JANNATUR ANJU"},{"id":19,"text":"RIYAD "},{"id":20,"text":"RAJIBUL HOQUE"},{"id":21,"text":"UMME SALMA"},{"id":22,"text":"RAFIKUL ISLAM"},{"id":23,"text":"ABJAL HUSSAIN"},{"id":24,"text":"IMAM AHMED"},{"id":25,"text":"NARGIS SULTANA"},{"id":26,"text":"MAFUDA KHATUN"},{"id":27,"text":"SIDDIKUL NAHAR"},{"id":28,"text":"KHALEDA BEGUM"}]

但如果我搜索,比如EJAJ,它说没有找到结果。我正在使用select版本4.0.3

摆脱整个
过程结果
的东西,让数据服务器将JSON数组变成一个子数组,其中包含键
结果
。这样:

{"results":[your-JSON-array-above]}

在processResults上,对数据进行计数,如果其长度==0,则执行smt

processResults: function (data) {
    if(data.length==0){
        getFromGoogle();
    }
    return {
        results: data
    };
},
processResults: function (data) {
    if(data.length==0){
        getFromGoogle();
    }
    return {
        results: data
    };
},