php - mysql_fetch_row() expects parameter 1 to be resource, object given.? -

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संभव डुप्लिकेट:

मेरे कोड में त्रुटि का कारण क्या है और मैं कैसे तय कर सकता हूं? (मैं कोडिगनिटर का उपयोग करता हूं) 

  $ query_hotel_search = $ this- & gt; db- & gt; क्वेरी ("SELECT * होटल के होटल से जहां नाम '% $ hotel_search%' नाम से नाम '' नाम है); $ इकाइयों = सरणी (); जबकि ($ row = mysql_fetch_row ($ query_hotel_search)) / / लाइन 27 {$ इकाइयों = unserialize ($ पंक्ति [0]); } गूंज json_encode (var_dump ($ इकाइयों));   
 त्रुटि:  
 
 एक PHP त्रुटि का सामना करना पड़ा था 
 गंभीरता: चेतावनी 
 संदेश: mysql_fetch_row () पैरामीटर 1 संसाधन होने की उम्मीद है, ऑब्जेक्ट दिए गए पंक्ति संख्या: 27 
   
 आउटपुट:  सरणी (0) {} शून्य   
  अद्यतन करें:   
 त्रुटि:  
 
 एक PHP त्रुटि का सामना करना पड़ा था 
 गंभीरता: सूचना / संदेश: unserialize () [function.unserialize]: त्रुटि ऑफ़सेट 0 3 बाइट्स - रेखा संख्या: 29 
   
 आउटपुट:  bool (false) null   
 मेरे डेटाबेस देखें :  
  $ query_hotel_search = mysql_query ("चयन करें * से होटल के नाम जहां '% $ hotel_search%' नाम से नाम दर्ज करें");; अगर (mysql_num_rows ($ query_hotel_search) == 0) {वापसी '0'; } और {$ इकाइयों = सरणी (); जबकि ($ पंक्ति = mysql_fetch_row ($ query_hotel_search)) {$ इकाइयों = अनसारीकृत ($ पंक्ति- & gt; इकाइयां [0]); // पंक्ति 2 9) इको जेसन_एएनडीड (var_dump ($ इकाइयां)); }    
 
 
  
 mysql_fetch_row () क्वेरी खुद नहीं लेता है लेकिन बजाय mysql_query कमांड के परिणाम (जो कि एक संसाधन है)।  
 PHP प्रलेखन (शीर्ष पर लिंक) पर उदाहरण देखें और आप mysql_fetch_row की कार्यक्षमता प्राप्त करेंगे।  
 > शायद $ query_hotel_search का असाइनमेंट स्थिति को स्पष्ट कर सकता है।

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