Question

我已在控制器中编写代码以将上传的文件移动到文件夹，但它无法正常工作，我的代码是

 $this->layout = "ajax";
    if($this->request->data){
    $postArr = $this->request->data;
    $a = $postArr['image'];
        $ext = substr(strtolower(strrchr($a, '.')), 1); //get the extension
                    $arr_ext = array('jpg', 'jpeg', 'gif'); //set allowed extensions
                   if(in_array($ext, $arr_ext))
                    {
                        //do the actual uploading of the file. First arg is the tmp name, second arg is
                        //where we are putting it
                      if(move_uploaded_file($a, WWW_ROOT . '/img/uploads/users' . $a)){
                      echo 'success';
                        $hai = 1;
                      }
                      else
                      {
                      echo 'failed';
                      $hai = 0;
                      }



                    }
                    .....
                    .....

                    ......
 $Curriculum = array();
 $Curriculum['Shop']['image'] = $hai;
        $this->Shop->save($Curriculum);
         echo 0;
        exit;

我认为这行'if（move_uploaded_file（$ a，WWW_ROOT。'/ img / uploads / users'。$ a））{...}无效，通过此代码，我可以存储'0在我的数据库中，也就是代码的一部分。

如何将上传的图像移动到文件夹和路径到我的数据库字段..plz帮我任何一个......！

我的脚本代码是

 function addAdminList(){
    var branchId = $('#branchId').val();
    var branchName = $('#branchName').val();
    var curiculumName = $('#curiculumName').val();
         var owner = $('#owner').val();
         var startdate = $('#startdate').val();
         var phone = $('#phone').val();
         var desc = $('#desc').val();
         var address = $('#address').val();
         var timings = $('#timings').val();
        var image = $('#image').val();                                   

    if(!$("#addClsGrpFrm").validationEngine('validate')){
        return false;
    }
    $('.overlay').show();
    $.ajax({
        url : '<?php echo BASE_PATH; ?>adminAjax/addAdminList',
        type : 'POST',
        data : {
            branchId : branchId,
            branchName : branchName,
            curiculumName : curiculumName,
            owner : owner,
            startdate : startdate,
            phone : phone,
            desc : desc,
            address : address,
            timings : timings,
            image : image
             },
        success : function(res){
            $('.overlay').hide();
            if(res == 0){

            }
            parent.$.colorbox.close();
            oTable.fnDraw();
            return false;
        },
        error : function(res){
            alert('Server Error Occurred'); 
        }
    });

}

我的表格是

    <form id="addClsGrpFrm" method="post" action="#" onsubmit="return disableFrmSubmit()">
        <div class="flash_msg"><?php echo $this->Session->flash(); ?></div>
        <input type="hidden" name="data[branch][id]" readonly id="branchId" value="0">
        <div class="login-form">
            <label>Shop Image</label>
            <input type="file" id="image" data-prompt-position="topLeft:200" name="data[branch][image]" placeholder="" class="validate[required]" />

   .....
    </div>
                <div class="login-form">
                    <input type="button" onclick="addAdminList()" class="login-btn" value="Submit">
                </div>
        <div class="clear"></div>
    </form>

Answer 1

提交表单时的一些建议：

在提交表单时使用 FormData ，而不是逐个添加。
您在控制器中使用的扩展提取过程可能无法使用“。”保存某些文件。在他们的文件名中。

我为您的知识创建了一个测试表。这是表格：

<form id="testForm" method="post" action="<?php echo 'http://localhost/blog/users/test'; ?>">
    <input type="file" name="image" id="image">
    <input type="submit" value="Submit">
</form>

您可以像这样使用FormData：

$('#testForm').on('submit', function(event) {
    event.preventDefault();
    var form_data = new FormData($(this)[0]); // this will get all your form's inputs
    $.ajax({
        url:"<?php echo 'http://localhost/blog/users/test.json'; ?>", // you can use your form action URL here
        type:'POST',
        data:form_data,
        processData: false,
        contentType: false
    });
});

最后在你的控制器中：

public function test() {
    if ($this->request->is('post')) {
        if (!empty($this->request->data['image']['tmp_name']) && is_uploaded_file($this->request->data['image']['tmp_name'])) {
            $filename = basename($this->request->data['image']['name']); 
            $imagePath = WWW_ROOT . DS . 'documents' . DS . $filename; // save the file where you wish
            move_uploaded_file(
                $this->request->data['image']['tmp_name'],
                $imagePath // save $imagePath in your database
            );
        }
    }
}

您可以使用$ imagePath保存在数据库中，对于文件扩展名，您可以使用此函数，因为即使文件名中包含“点”，它也会获取文件类型：

public function findFileTypeAndName($filename)
{
    $split_point = '.';  
    $parts = explode($split_point, $filename);
    $last = array_pop($parts);
    $fileNameAndType = array(implode($split_point, $parts), $last);

    return $fileNameAndType; 
}

Answer 2

 $this->layout = "ajax";
    if($this->request->data){
    $postArr = $this->request->data;
    $a = $postArr['image'];
        $ext = substr(strtolower(strrchr($a['name'], '.')), 1); //get the extension
                    $arr_ext = array('jpg', 'jpeg', 'gif'); //set allowed extensions
                   if(in_array($ext, $arr_ext))
                    {
                        //do the actual uploading of the file. First arg is the tmp name, second arg is
                        //where we are putting it
                      if(move_uploaded_file($a['tmp_name'], WWW_ROOT . '/img/uploads/users/' . basename($a['name']))){
                      echo 'success';
                        $hai = 1;
                      }
                      else
                      {
                      echo 'failed';
                      $hai = 0;
                      }



                    }
                    .....
                    .....

                    ......
 $Curriculum = array();
 $Curriculum['Shop']['image'] = $hai;
        $this->Shop->save($Curriculum);
         echo 0;
        exit;

问题只是您此时尝试将图像数组传递到strrchr($a)而不是strrchr($a['name'])：

$ext = substr(strtolower(strrchr($a['name'], '.')), 1); //get the extension

然后您还尝试将$a数组传递给move_uploaded_file($a,...)。检查我对代码部分的更正......

如何将图像上传到文件夹和路径ot db在cakephp控制器

2 个答案: