2014-02-22 33 views
3

我有不同的PHP變量包含我想放在mysql數據庫中的表中的一個數據。問題是我需要把內容放在一個特定的格式中,所以連接到數據庫的另一個也可以使用它。 我不知道這是什麼類型的格式。它看起來不像JSON是這個json以及如何用php變量替換值?

a:1:{i:0;s:27:"[email protected]";} 



a:17:{i:0;a:7:{s:2:"id";s:1:"1";s:4:"slug";s:19:"contact-information";s:4:"name";s:19:"Contact Information";s:4:"type";s:8:"fieldset";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:0:"";}i:1;a:7:{s:2:"id";s:1:"6";s:4:"slug";s:12:"company-name";s:4:"name";s:12:"Company Name";s:4:"type";s:4:"text";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:20:"Company Name is here";}i:2;a:7:{s:2:"id";s:1:"8";s:4:"slug";s:12:"contact-name";s:4:"name";s:12:"Contact Name";s:4:"type";s:4:"text";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:8:"Name";}i:3;a:7:{s:2:"id";s:2:"10";s:4:"slug";s:13:"email-address";s:4:"name";s:13:"Email Address";s:4:"type";s:5:"email";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:26:"[email protected]";}i:4;a:7:{s:2:"id";s:2:"12";s:4:"slug";s:15:"website-address";s:4:"name";s:15:"Website Address";s:4:"type";s:3:"url";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:25:"http://example.com";}i:5;a:7:{s:2:"id";s:2:"15";s:4:"slug";s:13:"contact-phone";s:4:"name";s:13:"Contact Phone";s:4:"type";s:5:"phone";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:12:"923246573632";}i:6;a:7:{s:2:"id";s:2:"16";s:4:"slug";s:12:"your-address";s:4:"name";s:12:"Your Address";s:4:"type";s:7:"address";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:91:"House # 2, Street 27<br>Chah Mrain<br>Lahore, Punjab 54000<br>New Zealand";}i:7;a:7:{s:2:"id";s:2:"27";s:4:"slug";s:19:"listing-information";s:4:"name";s:19:"Listing Information";s:4:"type";s:8:"fieldset";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:0:"";}i:8;a:7:{s:2:"id";s:2:"17";s:4:"slug";s:13:"listing-title";s:4:"name";s:13:"Listing Title";s:4:"type";s:4:"text";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:17:"House of the Dead";}i:9;a:7:{s:2:"id";s:2:"19";s:4:"slug";s:12:"listing-type";s:4:"name";s:12:"Listing Type";s:4:"type";s:6:"select";s:7:"options";s:90:"a:4:{i:0;s:11:"Attractions";i:1;s:13:"Accommodation";i:2;s:5:"Tours";i:3;s:9:"Transport";}";s:9:"parent_id";s:1:"0";s:5:"value";s:13:"Accommodation";}i:10;a:7:{s:2:"id";s:2:"20";s:4:"slug";s:17:"listings-overview";s:4:"name";s:19:"Listing\'s Overview";s:4:"type";s:8:"textarea";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:55:"Here is the overview of listing lets check what happens";}i:11;a:7:{s:2:"id";s:2:"21";s:4:"slug";s:10:"logo-image";s:4:"name";s:10:"Logo Image";s:4:"type";s:11:"file-upload";s:7:"options";s:31:"a:1:{i:0;s:13:"png|jpe?g|gif";}";s:9:"parent_id";s:1:"0";s:5:"value";s:80:"http://example.com/connect-icon.png";}i:12;a:7:{s:2:"id";s:2:"25";s:4:"slug";s:9:"longitude";s:4:"name";s:9:"Longitude";s:4:"type";s:4:"text";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:0:"";}i:13;a:7:{s:2:"id";s:2:"26";s:4:"slug";s:8:"latitude";s:4:"name";s:8:"Latitude";s:4:"type";s:4:"text";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:0:"";}i:14;a:7:{s:2:"id";s:1:"2";s:4:"slug";s:12:"verification";s:4:"name";s:12:"Verification";s:4:"type";s:12:"verification";s:7:"options";s:0:"";s:9:"parent_id";s:1:"0";s:5:"value";s:0:"";}i:15;a:7:{s:2:"id";s:1:"3";s:4:"slug";s:27:"please-enter-any-two-digits";s:4:"name";s:27:"Please enter any two digits";s:4:"type";s:6:"secret";s:7:"options";s:0:"";s:9:"parent_id";s:1:"2";s:5:"value";s:2:"14";}i:16;a:7:{s:2:"id";s:1:"4";s:4:"slug";s:6:"submit";s:4:"name";s:6:"Submit";s:4:"type";s:6:"submit";s:7:"options";s:0:"";s:9:"parent_id";s:1:"2";s:5:"value";s:0:"";}} 

我只想替換變量中的值並將其保存在數據庫中。

+0

轉到[jsonlint.com](jsonlint.com)並將此json代碼粘貼到文本區域並進行驗證,它會顯示您是否有效。我檢查與你的JSON它是無效的 –

回答

7

該格式是PHP的serialize()函數的結果。你可以使用unserialize()來解碼它。

+0

謝謝@Barmar - 我在這裏檢查了這裏http://www.functions-online.com/unserialize.html和是的,這是串行 – Muhammad

+0

謝謝巴爾馬,我試圖找到格式。它似乎是JSON,但它不是:) –