Question

我有以下PHP函数，如果'revision_no'字段不匹配，则将JSON对象插入mongodb，如果匹配，则更新这两个字段。

function saveJson($data){
    $mongo = new Mongo();
    $db = $mongo->selectDB("technical");
    $db->createCollection("ge",false);
    $technical = $db->selectCollection("ge");
    $json = json_decode($data['data']);
    $count = $technical->count(array('job_id' => $data['job_id'], 'revision_no' => $data['revision_no']));
    echo $count;
    if($count > 0){
        //With just job_id, it returns the correct count
        //$technical->update(array('job_id' => $data['job_id']), $json);
        $technical->update(array('job_id' => $data['job_id'], 'revision_no' => $data['revision_no']), $json);
    }else{
        $technical->insert($json);
    }
}

问题是我无法将计数与job_id＆amp;组合使用。 revision_no;但它确实只与job_id一起使用。谁能告诉我哪里出错了？

Answer 1

当我应该使用对象样式时，原来我正在使用数组语法！所以它不是$ data [＆＃39; revision_no＆＃39;]而是$ json-＆gt; revision_no

function saveJson($data){
    $mongo = new Mongo();
    $db = $mongo->selectDB("technical");
    $db->createCollection("ge",false);
    $technical = $db->selectCollection("ge");
    $json = json_decode($data['data']);
    $count = $technical->count(array('job_id' => $data['job_id'], 'revision_no' => $json->revision_no));
    echo $count;
    if($count > 0){
        //With just job_id, it returns the correct count
        //$technical->update(array('job_id' => $data['job_id']), $json);
        $technical->update(array('job_id' => $data['job_id'], 'revision_no' => $json->revision_no), $json);
    }else{
        $technical->insert($json);
    }
}

PHP Mongo计数不适用于多个字段

1 个答案: