Home >Backend Development >PHP Tutorial >How to get the binary data of the image uploaded by formData in php

How to get the binary data of the image uploaded by formData in php

WBOY
WBOYOriginal
2016-09-14 09:41:232286browse

<code>                        <input id="file" type="file" accept="image/png,image/gif,image/jpeg" name="file">
                        ----------------------------------------------------------------
                        
                        var fileInput = document.getElementById("file");
                        var file = fileInput.files[0];
                        var formData = new FormData();
                        formData.append("file", file);

                        $.ajax({
                            url: "./upload_photobank.php",
                            type: "POST",
                            data: formData,
                            processData: false,  // 告诉jQuery不要去处理发送的数据
                            contentType: false,   // 告诉jQuery不要去设置Content-Type请求头
                            complete : function(jqXHR, textStatus) {
                                if(jqXHR.status != 200){
                                    console.log( 456 )
                                }else{
                                    var jsonData = eval('(' + jqXHR.responseText + ')');
                                    // var jsonData = jqXHR.responseText;                                        
                                    console.log(jqXHR.responseText);
                                }
                            }
                        });</code>

How to write PHP to obtain the binary data of uploaded images?
Are there any errors in the js part?

Reply content:

<code>                        <input id="file" type="file" accept="image/png,image/gif,image/jpeg" name="file">
                        ----------------------------------------------------------------
                        
                        var fileInput = document.getElementById("file");
                        var file = fileInput.files[0];
                        var formData = new FormData();
                        formData.append("file", file);

                        $.ajax({
                            url: "./upload_photobank.php",
                            type: "POST",
                            data: formData,
                            processData: false,  // 告诉jQuery不要去处理发送的数据
                            contentType: false,   // 告诉jQuery不要去设置Content-Type请求头
                            complete : function(jqXHR, textStatus) {
                                if(jqXHR.status != 200){
                                    console.log( 456 )
                                }else{
                                    var jsonData = eval('(' + jqXHR.responseText + ')');
                                    // var jsonData = jqXHR.responseText;                                        
                                    console.log(jqXHR.responseText);
                                }
                            }
                        });</code>

How to write PHP to obtain the binary data of uploaded images?
Are there any errors in the js part?

First make sure that the attributes of the form must have enctype="multipart/form-data", and the instantiated FormData cannot be empty! If it is empty, the form value cannot be obtained

Use $_FILES instead of $_POST

The upload code must be written into the "change" event of the input... and the ajax callback is a bit strange. Once it is compiled, don't judge the status anymore. .
You can refer to the front-end picture I wrote for direct transmission to OSS test

FormData is a form. All data processing is the same as the form, so PHP also processes it the same way. Uploaded files are still processed through $_FILES.

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn