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PHP table_sql::query_db方法代码示例

本文整理汇总了PHP中table_sql::query_db方法的典型用法代码示例。如果您正苦于以下问题:PHP table_sql::query_db方法的具体用法?PHP table_sql::query_db怎么用?PHP table_sql::query_db使用的例子?那么, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在table_sql的用法示例。


在下文中一共展示了table_sql::query_db方法的3个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的PHP代码示例。

示例1: query_db

    public function query_db($pagesize, $useinitialsbar = true) {
        $doneslots = array();
        foreach ($this->get_sort_columns() as $column => $notused) {
            $slot = $this->is_latest_step_column($column);
            if ($slot && !in_array($slot, $doneslots)) {
                $this->add_latest_state_join($slot);
                $doneslots[] = $slot;
            }
        }

        parent::query_db($pagesize, $useinitialsbar);

        if ($this->requires_latest_steps_loaded()) {
            $qubaids = $this->get_qubaids_condition();
            $this->lateststeps = $this->load_question_latest_steps($qubaids);
        }
    }
开发者ID:Jtgadbois,项目名称:Pedadida,代码行数:17,代码来源:attemptsreport_table.php

示例2: explode

 function query_db($pagesize, $useinitialsbar = true)
 {
     // Add table joins so we can sort by question answer
     // unfortunately can't join all tables necessary to fetch all answers
     // to get the state for one question per attempt row we must join two tables
     // and there is a limit to how many joins you can have in one query. In MySQL it
     // is 61. This means that when having more than 29 questions the query will fail.
     // So we join just the tables needed to sort the attempts.
     if ($sort = $this->get_sql_sort()) {
         $this->sql->from .= ' ';
         $sortparts = explode(',', $sort);
         $matches = array();
         foreach ($sortparts as $sortpart) {
             $sortpart = trim($sortpart);
             if (preg_match('/^qsanswer([0-9]+)/', $sortpart, $matches)) {
                 $qid = intval($matches[1]);
                 $this->sql->fields .= ", qs{$qid}.grade AS qsgrade{$qid}, qs{$qid}.answer AS qsanswer{$qid}, qs{$qid}.event AS qsevent{$qid}, qs{$qid}.id AS qsid{$qid}";
                 $this->sql->from .= "LEFT JOIN {question_sessions} qns{$qid} ON qns{$qid}.attemptid = qa.uniqueid AND qns{$qid}.questionid = :qid{$qid} ";
                 $this->sql->from .= "LEFT JOIN  {question_states} qs{$qid} ON qs{$qid}.id = qns{$qid}.newgraded ";
                 $this->sql->params['qid' . $qid] = $qid;
             }
         }
     }
     parent::query_db($pagesize, $useinitialsbar);
     $qsfields = 'qs.id, qs.grade, qs.event, qs.question, qs.answer, qs.attempt';
     if (!$this->is_downloading()) {
         $attemptids = array();
         foreach ($this->rawdata as $attempt) {
             if ($attempt->uniqueid > 0) {
                 $attemptids[] = $attempt->uniqueid;
             }
         }
         $this->gradedstatesbyattempt = quiz_get_newgraded_states($attemptids, true, $qsfields);
     } else {
         $this->gradedstatesbyattempt = quiz_get_newgraded_states($this->sql, true, $qsfields);
     }
 }
开发者ID:ajv,项目名称:Offline-Caching,代码行数:37,代码来源:responses_table.php

示例3: explode

 function query_db($pagesize, $useinitialsbar = true)
 {
     // Add table joins so we can sort by question grade
     // unfortunately can't join all tables necessary to fetch all grades
     // to get the state for one question per attempt row we must join two tables
     // and there is a limit to how many joins you can have in one query. In MySQL it
     // is 61. This means that when having more than 29 questions the query will fail.
     // So we join just the tables needed to sort the attempts.
     if ($sort = $this->get_sql_sort()) {
         if ($this->detailedmarks) {
             $this->sql->from .= ' ';
             $sortparts = explode(',', $sort);
             $matches = array();
             foreach ($sortparts as $sortpart) {
                 $sortpart = trim($sortpart);
                 if (preg_match('/^qsgrade([0-9]+)/', $sortpart, $matches)) {
                     $qid = intval($matches[1]);
                     $this->sql->fields .= ", qs{$qid}.grade AS qsgrade{$qid}, qs{$qid}.event AS qsevent{$qid}, qs{$qid}.id AS qsid{$qid}";
                     $this->sql->from .= "LEFT JOIN {question_sessions} qns{$qid} ON qns{$qid}.attemptid = qa.uniqueid AND qns{$qid}.questionid = :qid{$qid} ";
                     $this->sql->from .= "LEFT JOIN  {question_states} qs{$qid} ON qs{$qid}.id = qns{$qid}.newgraded ";
                     $this->sql->params['qid' . $qid] = $qid;
                 }
             }
         } else {
             //unset any sort columns that sort on question grade as the
             //grades are not being fetched as fields
             $sess =& $this->sess;
             foreach ($sess->sortby as $column => $order) {
                 if (preg_match('/^qsgrade([0-9]+)/', trim($column))) {
                     unset($sess->sortby[$column]);
                 }
             }
         }
     }
     parent::query_db($pagesize, $useinitialsbar);
     //get all the attempt ids we want to display on this page
     //or to export for download.
     if (!$this->is_downloading()) {
         $attemptids = array();
         foreach ($this->rawdata as $attempt) {
             if ($attempt->attemptuniqueid > 0) {
                 $attemptids[] = $attempt->attemptuniqueid;
             }
         }
         $this->gradedstatesbyattempt = quiz_get_newgraded_states($attemptids, true, 'qs.id, qs.grade, qs.event, qs.question, qs.attempt');
         if (has_capability('mod/quiz:regrade', $this->context)) {
             $this->regradedqs = quiz_get_regraded_qs($attemptids);
         }
     } else {
         $this->gradedstatesbyattempt = quiz_get_newgraded_states($this->sql, true, 'qs.id, qs.grade, qs.event, qs.question, qs.attempt');
         if (has_capability('mod/quiz:regrade', $this->context)) {
             $this->regradedqs = quiz_get_regraded_qs($this->sql);
         }
     }
 }
开发者ID:nicolasconnault,项目名称:moodle2.0,代码行数:55,代码来源:overview_table.php


注:本文中的table_sql::query_db方法示例由纯净天空整理自Github/MSDocs等开源代码及文档管理平台,相关代码片段筛选自各路编程大神贡献的开源项目,源码版权归原作者所有,传播和使用请参考对应项目的License;未经允许,请勿转载。