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<code>bug:SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data. $("#file-5").fileinput({ language: 'zh', //设置语言 uploadUrl: "{:U('profile/upload')}", // avatar_uploadyou must set a valid URL here else you will get an error __UPLOAD__partner/img allowedFileExtensions : ['jpg', 'png','gif'], overwriteInitial: false, maxFileSize: 1000, maxFilesNum: 1, //allowedFileTypes: ['image', 'video', 'flash'], slugCallback: function(filename) { alert("aaa"); return filename.replace('(', '_').replace(']', '_'); } }) .on("fileuploaded", function (e, data) { var res = data.response; if (res.state > 0) { alert('上传成功'); alert(res.path); } else { alert('上传失败') } }); <form enctype="multipart/form-data"> <hr> <div class="form-group"> <input id="file-5" class="file" type="file" multiple data-preview-file-type="any" data-upload-url="#" name="image_data"> </div> </form> 后台程序我有public function upload(){ }方法,是php的 </code>
The screenshot of the error seems to be a syntax error. I suspect that there is an error in the fileinput event in the foreground. Because I only return one value in the background for testing, it should be fine. I am not familiar with fileinput. Please help! ! ! If anyone has an example for me to refer to, thank you!
<code>bug:SyntaxError: JSON.parse: unexpected character at line 1 column 1 of the JSON data. $("#file-5").fileinput({ language: 'zh', //设置语言 uploadUrl: "{:U('profile/upload')}", // avatar_uploadyou must set a valid URL here else you will get an error __UPLOAD__partner/img allowedFileExtensions : ['jpg', 'png','gif'], overwriteInitial: false, maxFileSize: 1000, maxFilesNum: 1, //allowedFileTypes: ['image', 'video', 'flash'], slugCallback: function(filename) { alert("aaa"); return filename.replace('(', '_').replace(']', '_'); } }) .on("fileuploaded", function (e, data) { var res = data.response; if (res.state > 0) { alert('上传成功'); alert(res.path); } else { alert('上传失败') } }); <form enctype="multipart/form-data"> <hr> <div class="form-group"> <input id="file-5" class="file" type="file" multiple data-preview-file-type="any" data-upload-url="#" name="image_data"> </div> </form> 后台程序我有public function upload(){ }方法,是php的 </code>
The screenshot of the error seems to be a syntax error. I suspect that there is an error in the fileinput event in the foreground. Because I only return one value in the background for testing, it should be fine. I am not familiar with fileinput. Please help! ! ! If anyone has an example for me to refer to, thank you!