Question

如何使用代码点火器

读取review_text列值

在这行代码中，igniter给出了以下错误

<p class="excerpt" align="justify"><?=$review->REVIEW_TEXT;?></p>

在网页上显示时给出错误

A PHP Error was encountered

Severity: Notice

Message: Trying to get property of non-object

Filename: views/reviewpage.php

Line Number: 83

这是我的表

CREATE TABLE  `techreview`.`trn_reviews` (
  `review_id` int(10) unsigned NOT NULL auto_increment,
  `review_url` varchar(200) collate latin1_general_ci NOT NULL,
  `review_text` text collate latin1_general_ci NOT NULL,
  `review_dt` date NOT NULL,
  PRIMARY KEY  (`review_id`)
) ENGINE=MyISAM DEFAULT CHARSET=latin1 COLLATE=latin1_general_ci;

这是我的模型方法

function getReviewDetails($url){    
    $this->db->select('REVIEW_URL,REVIEW_TEXT');
    $whereCondition = $array = array('REVIEW_URL' =>$url);       
    $this->db->where($whereCondition); 
    $query = $this->db->get("trn_reviews");
    return $query->result();            
}

Answer 1

我假设你正在做这样的事情？

<?php
  $review = $this->my_model->getReviewDetails("url_text_here");
?>
<p class="excerpt" align="justify"><?=$review->REVIEW_TEXT;?></p>

你不应该遍历结果集吗？

<?php
  $review = $this->my_model->getReviewDetails("url_text_here");
  foreach($review as $row) {
?>
<p class="excerpt" align="justify"><?=$row->review_text;?></p>
<?php
  }
?>

如果您的查询只返回1行，则可以使用$query->row()方法：

// in your model
function getReviewDetails($url){    
    $this->db->select('REVIEW_URL,REVIEW_TEXT');
    $whereCondition = $array = array('REVIEW_URL' =>$url);       
    $this->db->where($whereCondition); 
    $query = $this->db->get("trn_reviews");
    return $query->row();          
}

// in your view
<?php
  $review = $this->my_model->getReviewDetails("url_text_here");
?>
<p class="excerpt" align="justify"><?=$review->review_text;?></p>

http://codeigniter.com/user_guide/database/results.html

在显示TEXT类型列值时尝试获取非目标代码点火器的属性

1 个答案: