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使用codeigniter将row

2025-07-17 02:06:35
使用codeigniter将row_array添加到result_array foreach循环中?(Adding row_array to result_array foreach loop with codeigniter?) 我目前在一个网络应用程序上有一个功能,我正在建立一个求职者可以查看他们申请的工作,非常简单。 当它们“应用”时,该应用程序存储
使用codeigniter将row_array添加到result_array foreach循环中?(Adding row_array to result_array foreach loop with codeigniter?)

我目前在一个网络应用程序上有一个功能,我正在建立一个求职者可以查看他们申请的工作,非常简单。

当它们“应用”时,该应用程序存储在数据库表'applicati'中,并带有'job_id'列,该列存储来自'jobs'数据库表的作业的'id'。

目前,我能够拉出所述求职者所做的每个应用程序。

但是,我无法遍历每个应用程序并到与该应用程序对应的作业,然后将row_array()添加到更大的数组中,然后我将使用foreach循环输出作业。

基本上我问我如何将一个数组添加到一个数组,然后输出完整的数组?

appliedfor.php (COTROLLER)

$applicati_query = $this->db->get_where( 'applicati', array('jobseeker_profile_id' => $user['id']) ); $applicati = $applicati_query->result_array(); $data[] = array(); foreach ($applicati as $application) { $job_id = $application['job_id']; $data['job'] = $this->db->get_where('jobs', array('id' => $job_id))->row_array(); $data['jobs'] .= $data['job']; } $data['jobs']; $this->load->view('header'); $this->load->view('appliedfor', $data);

appliedfor.php (VIEW)

foreach ($jobs as $job) { $single_job_id = $job['id']; echo "<br>"; echo form_open('job/view' . '" id="eachJob'); echo "<div id=\"leftContain\" class=\"floatLeft\">"; echo "<h4 class=\"green\">" . $job['role'] . "</h4>"; echo "<div class=\"italic\"><div class=\"blue floatLeft\">" . $job['company'] . " &nbsp; </div><div class=\"floatLeft\">in</div><div class=\"blue floatLeft\"> &nbsp; " . $job['location'] . "</div></div><br><br>"; echo "</div>"; echo "<div id=\"rightContain\" class=\"floatLeft\">"; echo "<input type=\"hidden\" name=\"job_id\" value=\"" . $single_job_id . "\">"; echo form_submit('submit' . '" class="jobButton floatRight"', 'View Job'); echo "</div>"; echo form_close(); }

我目前得到2个错误:未定义的索引:作业和错误在这一行显然是foreach中控制器中的$data['jobs'] 。

另一个错误是视图文件中的foreach,但这基本上是由第一个错误触发的。

谢谢你的帮助。

I currently have a function on a web app which I'm building where a jobseeker can view the jobs they have applied for, very simple.

When they "apply" this application is stored in the database table 'applicati' with a 'job_id' column which stores the 'id' of the job from the 'jobs' database table.

At the moment I am able to pull each application the said jobseeker has made.

Though, I am unable to loop through each application and find the job which corresponds to that application and then add the row_array() to larger array which I will then output the jobs with a foreach loop.

Essentially I am asking how do I add an array to an array and then output the full array?

appliedfor.php (COTROLLER)

$applicati_query = $this->db->get_where( 'applicati', array('jobseeker_profile_id' => $user['id']) ); $applicati = $applicati_query->result_array(); $data[] = array(); foreach ($applicati as $application) { $job_id = $application['job_id']; $data['job'] = $this->db->get_where('jobs', array('id' => $job_id))->row_array(); $data['jobs'] .= $data['job']; } $data['jobs']; $this->load->view('header'); $this->load->view('appliedfor', $data);

appliedfor.php (VIEW)

foreach ($jobs as $job) { $single_job_id = $job['id']; echo "<br>"; echo form_open('job/view' . '" id="eachJob'); echo "<div id=\"leftContain\" class=\"floatLeft\">"; echo "<h4 class=\"green\">" . $job['role'] . "</h4>"; echo "<div class=\"italic\"><div class=\"blue floatLeft\">" . $job['company'] . " &nbsp; </div><div class=\"floatLeft\">in</div><div class=\"blue floatLeft\"> &nbsp; " . $job['location'] . "</div></div><br><br>"; echo "</div>"; echo "<div id=\"rightContain\" class=\"floatLeft\">"; echo "<input type=\"hidden\" name=\"job_id\" value=\"" . $single_job_id . "\">"; echo form_submit('submit' . '" class="jobButton floatRight"', 'View Job'); echo "</div>"; echo form_close(); }

I am currently getting 2 errors: Undefined index: jobs and the error is on this line apparently $data['jobs'] in the controller within the foreach.

The other error is the foreach within the view file but that is basically triggered by the first error.

Thanks for your help.

最满意答案

你是对的:

$data['jobs'] .= $data['job'];

连接字符串,而不是数组,永远不会工作。

相反,尝试类似于:

$data['jobs'][] = $data['job'];

构建一个作业数组,然后使用foreach()或任何最合适的输出

You are correct:

$data['jobs'] .= $data['job'];

concatenates strings, not arrays, and will never work.

Instead, try something like:

$data['jobs'][] = $data['job'];

to build up an array of the jobs, then output with a foreach() or whatever is most suitable

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10分钟前 发表
然后使用foreach()或任何最合适的输出 You are correct
本站网友 回收公司
12分钟前 发表
$data['jobs'] .= $data['job']; concatenates strings
本站网友 顺差
9分钟前 发表
and will never work. Instead
本站网友 黑云压城城欲摧
2分钟前 发表
而不是数组